Question Statement

Q3. Solve the following trigonometric equations when $ϕ \in [0, 2 π]$ : (i) $6 sin ϕ + 8 cos ϕ = 7$ (ii) $cos ϕ + cos 3 ϕ = 0$ (iii) $sin 4 ϕ + sin 3 ϕ = 0$ (iv) $sin ϕ + sin 3 ϕ = sin 2 ϕ$ (v) $tan ϕ + cot ϕ = 8 cos 2 ϕ$ (vi) $sin 2 ϕ + sin 3 ϕ + sin 5 ϕ = 0$ (vii) $5 cos ϕ + 4 = 2 sin^{2} ϕ$ (viii) $cos 2 ϕ + 3 = 5 cos ϕ$ (ix) $sin^{2} ϕ = 3 cos^{2} ϕ$ (x) $cos^{2} ϕ - 2 cos ϕ + 1 = 0$ (xi) $tan^{2} ϕ - tan ϕ = 6$

Q4. Solve the following trigonometric equations when $x \in [- π, π]$ : (i) $3 cos x + sin x = 2$ (ii) $4 cos x + 3 sin x = 2$ (iii) $3 sin x - 2 cos x = 1$ (iv) $3 cos x + sin x = 2$ (v) $7 cos x - 6 sin x = 4$ (vi) $4 cos x + 2 sin x = 5$ (vii) $sec x + 5 tan x + 12 = 0$ (viii) $sin 2 x - cos x = 0$ (ix) $cos^{2} x - sin^{2} x = 1$ (x) $3 sin^{2} x - 2 sin x = 0$

Background and Explanation

Solving trigonometric equations requires reducing the expression to a single trigonometric ratio using fundamental identities (like $sin^{2} θ + cos^{2} θ = 1$ ) or sum-to-product formulas. Once simplified, we solve for the variable within a specific interval, often by finding a reference angle and determining the appropriate quadrants. For equations involving squaring, it is essential to check for extraneous solutions.

Solution

Q3. Solutions for $ϕ \in [0, 2 π]$

## Part (i)

$6 sin ϕ + 8 cos ϕ = 7$ Squaring both sides: $(6 sin ϕ + 8 cos ϕ)^{2} = (7)^{2}$ $36 sin^{2} ϕ + 64 cos^{2} ϕ + 96 sin ϕ cos ϕ = 49$ Dividing both sides by $cos^{2} ϕ$ : $36 tan^{2} ϕ + 64 + 96 tan ϕ = 49 sec^{2} ϕ$ $36 tan^{2} ϕ + 64 + 96 tan ϕ = 49 (1 + tan^{2} ϕ)$ $13 tan^{2} ϕ - 96 tan ϕ - 15 = 0$ Using the quadratic formula: $tan ϕ = \frac{96 \pm ( - 96 ) ^{2} - 4 ( 13 ) ( - 15 )}{2 \times 13} = \frac{96 \pm 14 51}{26} = \frac{48 \pm 7 51}{13}$ $tan ϕ \approx 7.537$ or $tan ϕ \approx - 0.153$ $ϕ = 82.4 4^{\circ}$ or $ϕ = - 8.70 3^{\circ}$ Converting to radians: $S . S = {\frac{82.44 π}{180}, \frac{- 8.703 π}{180}}$

## Part (ii)

$cos 3 ϕ + cos ϕ = 0$ Using $cos α + cos β = 2 cos (\frac{α + β}{2}) cos (\frac{α - β}{2})$ : $2 cos (2 ϕ) cos (ϕ) = 0$

$cos 2 ϕ = 0 \Rightarrow 2 ϕ = \frac{π}{2}, \frac{3 π}{2} \Rightarrow ϕ = \frac{π}{4}, \frac{3 π}{4}$
$cos ϕ = 0 \Rightarrow ϕ = \frac{π}{2}, \frac{3 π}{2}$ $S . S = {\frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}, \frac{3 π}{2}}$

## Part (iii)

$sin 4 ϕ + sin 3 ϕ = 0$ Using sum-to-product formula: $2 sin (\frac{7 ϕ}{2}) cos (\frac{ϕ}{2}) = 0$

$sin (\frac{7 ϕ}{2}) = 0 \Rightarrow \frac{7 ϕ}{2} = 0, π, 2 π \Rightarrow ϕ = 0, \frac{2 π}{7}, \frac{4 π}{7}$
$cos (\frac{ϕ}{2}) = 0 \Rightarrow \frac{ϕ}{2} = \frac{π}{2} \Rightarrow ϕ = π$ $S . S = {0, π, \frac{2 π}{7}, \frac{4 π}{7}}$

## Part (iv)

$sin 3 ϕ + sin ϕ = sin 2 ϕ$ $2 sin 2 ϕ cos ϕ = sin 2 ϕ$ $sin 2 ϕ (2 cos ϕ - 1) = 0$

$sin 2 ϕ = 0 \Rightarrow 2 ϕ = 0, π, 2 π \Rightarrow ϕ = 0, \frac{π}{2}, π$
$2 cos ϕ = 1 \Rightarrow cos ϕ = \frac{1}{2} \Rightarrow ϕ = \frac{π}{3}, \frac{5 π}{3}$ $S . S = {0, \frac{π}{2}, π, \frac{π}{3}, \frac{5 π}{3}}$

## Part (v)

$tan ϕ + cot ϕ = 8 cos 2 ϕ$ $\frac{s i n ^{2} ϕ + c o s ^{2} ϕ}{s i n ϕ c o s ϕ} = 8 cos 2 ϕ \Rightarrow \frac{1}{s i n ϕ c o s ϕ} = 8 cos 2 ϕ$ $1 = 4 (2 sin ϕ cos ϕ) cos 2 ϕ \Rightarrow 1 = 4 sin 2 ϕ cos 2 ϕ$ $1 = 2 sin 4 ϕ \Rightarrow sin 4 ϕ = \frac{1}{2}$ Reference angle is $\frac{π}{6}$ .

$4 ϕ = \frac{π}{6} \Rightarrow ϕ = \frac{π}{24}$
$4 ϕ = π - \frac{π}{6} = \frac{5 π}{6} \Rightarrow ϕ = \frac{5 π}{24}$ $S . S = {\frac{π}{24}, \frac{5 π}{24}}$

## Part (vi)

$sin 5 ϕ + sin 3 ϕ + sin 2 ϕ = 0$ $2 sin 4 ϕ cos ϕ + sin 2 ϕ = 0$ $2 (2 sin 2 ϕ cos 2 ϕ) cos ϕ + sin 2 ϕ = 0$ $sin 2 ϕ [4 cos 2 ϕ cos ϕ + 1] = 0$

$sin 2 ϕ = 0 \Rightarrow ϕ = 0, \frac{π}{2}, π$
$4 (2 cos^{2} ϕ - 1) cos ϕ + 1 = 0 \Rightarrow 8 cos^{3} ϕ - 4 cos ϕ + 1 = 0$ By inspection, $cos ϕ = \frac{1}{2}$ is a root. Using synthetic division: Depressed equation: $8 cos^{2} ϕ + 4 cos ϕ - 2 = 0$ .

From $cos ϕ = \frac{1}{2}$ , $ϕ = \frac{π}{3}, \frac{5 π}{3}$ .
From quadratic: $cos ϕ = \frac{- 1 \pm 5}{4}$ .
$cos ϕ = \frac{- 1 + 5}{4} \Rightarrow ϕ = 7 2^{\circ} = \frac{2 π}{5}$ .
$cos ϕ = \frac{- 1 - 5}{4} \Rightarrow ϕ = 14 4^{\circ} = \frac{4 π}{5}$ . $S . S = {0, \frac{π}{2}, π, \frac{π}{3}, \frac{5 π}{3}, \frac{2 π}{5}, \frac{4 π}{5}}$

## Part (vii)

$5 cos ϕ + 4 = 2 (1 - cos^{2} ϕ)$ $2 cos^{2} ϕ + 5 cos ϕ + 2 = 0$ $(2 cos ϕ + 1) (cos ϕ + 2) = 0$

$cos ϕ = - 2$ (Impossible)
$cos ϕ = - 1/2 \Rightarrow ϕ = π - \frac{π}{3}, π + \frac{π}{3} \Rightarrow ϕ = \frac{2 π}{3}, \frac{4 π}{3}$ $S . S = {\frac{2 π}{3}, \frac{4 π}{3}}$

## Part (viii)

$(2 cos^{2} ϕ - 1) + 3 = 5 cos ϕ$ $2 cos^{2} ϕ - 5 cos ϕ + 2 = 0$ $(2 cos ϕ - 1) (cos ϕ - 2) = 0$

$cos ϕ = 2$ (Impossible)
$cos ϕ = 1/2 \Rightarrow ϕ = \frac{π}{3}, \frac{5 π}{3}$ $S . S = {\frac{π}{3}, \frac{5 π}{3}}$

## Part (ix)

$sin^{2} ϕ = 3 cos^{2} ϕ \Rightarrow tan^{2} ϕ = 3$ $tan ϕ = \pm 3$ $ϕ = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}$ $S . S = {\frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}}$

## Part (x)

$(cos ϕ - 1)^{2} = 0 \Rightarrow cos ϕ = 1$ $ϕ = 0, 2 π$ $S . S = {0, 2 π}$

## Part (xi)

$tan^{2} ϕ - tan ϕ - 6 = 0$ $(tan ϕ - 3) (tan ϕ + 2) = 0$

$tan ϕ = 3 \Rightarrow ϕ = \frac{71.56 π}{180}, \frac{251.56 π}{180}$
$tan ϕ = - 2 \Rightarrow ϕ = \frac{116.57 π}{180}, \frac{296.57 π}{180}$ $S . S = {\frac{71.56 π}{180}, \frac{116.57 π}{180}, \frac{251.56 π}{180}, \frac{296.57 π}{180}}$

Q4. Solutions for $x \in [- π, π]$

## Part (i)

$3 cos x + sin x = 2 \Rightarrow 3 cos x = 2 - sin x$ Squaring: $9 (1 - sin^{2} x) = 4 + sin^{2} x - 4 sin x \Rightarrow 10 sin^{2} x - 4 sin x - 5 = 0$ $sin x = \frac{2 \pm 3 6}{10}$ .

$sin x \approx 0.9348 \Rightarrow x = 69. 2^{\circ}, 110. 8^{\circ}$ . Checking: $x = 69. 2^{\circ}$ satisfies.
$sin x \approx - 0.5348 \Rightarrow x = - 32.3 3^{\circ}, - 147.6 7^{\circ}$ . Checking: $x = - 32.3 3^{\circ}$ satisfies. $S . S = {\frac{- 32.33 π}{180}, \frac{69.2 π}{180}}$

## Part (ii)

$4 cos x + 3 sin x = 2 \Rightarrow 25 sin^{2} x - 12 sin x - 12 = 0$ $sin x = \frac{6 \pm 4 21}{25}$ .

$sin x \approx 0.9732 \Rightarrow x = 76. 7^{\circ}, 103. 3^{\circ}$ . Checking: $x = 103. 3^{\circ}$ satisfies.
$sin x \approx - 0.4932 \Rightarrow x = - 29.5 5^{\circ}, - 150.4 5^{\circ}$ . Checking: $x = - 29.5 5^{\circ}$ satisfies. $S . S = {\frac{- 29.55 π}{180}, \frac{103.3 π}{180}}$

## Part (iii)

$3 sin x - 2 cos x = 1 \Rightarrow 13 sin^{2} x - 6 sin x - 3 = 0$ $sin x = \frac{3 \pm 4 3}{13}$ .

$sin x \approx 0.76 \Rightarrow x = 49.4 6^{\circ}, 130.5 4^{\circ}$ . Checking: $x = 49.4 6^{\circ}$ satisfies.
$sin x \approx - 0.30 \Rightarrow x = - 17.4 6^{\circ}, - 162.5 4^{\circ}$ . Checking: $x = - 162.5 4^{\circ}$ satisfies. $S . S = {\frac{- 162.54 π}{180}, \frac{49.46 π}{180}}$

## Part (iv)

$3 cos x + sin x = 2 \Rightarrow 4 sin^{2} x - 22 sin x - 1 = 0$ $sin x = \frac{2 \pm 6}{4}$ .

$sin x \approx 0.9659 \Rightarrow x = 7 5^{\circ} (\frac{5 π}{12}), 10 5^{\circ}$ . Checking: $x = \frac{5 π}{12}$ satisfies.
$sin x \approx - 0.2588 \Rightarrow x = - 1 5^{\circ} (- \frac{π}{12}), - 16 5^{\circ}$ . Checking: $x = - \frac{π}{12}$ satisfies. $S . S = {\frac{- π}{12}, \frac{5 π}{12}}$

## Part (v)

$7 cos x - 6 sin x = 4 \Rightarrow 85 sin^{2} x + 48 sin x - 33 = 0$ $sin x = \frac{- 24 \pm 7 69}{85}$ .

$sin x \approx 0.4017 \Rightarrow x = 23.6 8^{\circ}$ . Checking: Satisfied.
$sin x \approx - 0.9664 \Rightarrow x = - 75.1 0^{\circ}, - 104. 9^{\circ}$ . Checking: Neither satisfies. $S . S = {\frac{23.68 π}{180}}$

## Part (vi)

$4 cos x + 2 sin x = 5 \Rightarrow 20 sin^{2} x - 45 sin x - 11 = 0$ $sin x = \frac{5 \pm 2 15}{10}$ .

$sin x \approx 0.9982 \Rightarrow x = 86.5 6^{\circ}$ . Checking: Satisfied.
$sin x \approx - 0.5509 \Rightarrow x = - 33.4 3^{\circ}$ . Checking: Satisfied. $S . S = {\frac{- 33.43 π}{180}, \frac{86.56 π}{180}}$

## Part (vii)

$sec x + 5 tan x + 12 = 0 \Rightarrow 169 sin^{2} x + 10 sin x - 143 = 0$ $sin x = \frac{- 5 \pm 24 42}{169}$ .

$sin x \approx 0.8907 \Rightarrow x = 117.0 4^{\circ}$ . Checking: Satisfied.
$sin x \approx - 0.9499 \Rightarrow x = - 71.7 8^{\circ}$ . Checking: Satisfied. $S . S = {\frac{- 71.78 π}{180}, \frac{117.04 π}{180}}$

## Part (viii)

$2 sin x cos x - cos x = 0 \Rightarrow cos x (2 sin x - 1) = 0$

$cos x = 0 \Rightarrow x = \frac{π}{2}$ (within domain)
$sin x = 1/2 \Rightarrow x = \frac{π}{6}, \frac{5 π}{6}$ $S . S = {\frac{π}{2}, \frac{π}{6}, \frac{5 π}{6}}$

## Part (ix)

$cos^{2} x - sin^{2} x = 1 \Rightarrow cos 2 x = 1$ $2 x = 0 \Rightarrow x = 0, \pm π$ $S . S = {0, \pm π}$

## Part (x)

$sin x (3 sin x - 2) = 0$

$sin x = 0 \Rightarrow x = 0, \pm π$
$sin x = 2/3 \Rightarrow x = 41.8 1^{\circ}, 138.1 9^{\circ}$ $S . S = {0, \pm π, \frac{41.81 π}{180}, \frac{138.19 π}{180}}$

Key Formulas or Methods Used

Pythagorean Identity: $sin^{2} θ + cos^{2} θ = 1$
Sum-to-Product Formulas:
- $cos α + cos β = 2 cos (\frac{α + β}{2}) cos (\frac{α - β}{2})$
- $sin α + sin β = 2 sin (\frac{α + β}{2}) cos (\frac{α - β}{2})$
Double Angle Formulas:
- $sin 2 θ = 2 sin θ cos θ$
- $cos 2 θ = 2 cos^{2} θ - 1 = 1 - 2 sin^{2} θ$
Quadratic Formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$
Synthetic Division: Used for solving higher-degree polynomial equations in terms of trig ratios.

Summary of Steps

Simplify the Equation: Use trigonometric identities to express the equation in terms of a single trigonometric function or a factorable form.
Solve Algebraically: Treat the trigonometric ratio as a variable (e.g., let $u = sin x$ ) and solve the resulting linear or quadratic equation.
Find Reference Angles: Use the inverse trigonometric function to find the base angle in the first quadrant.
Identify Quadrants: Based on the sign of the ratio, determine which quadrants the solutions lie in.
Apply Domain Constraints: Calculate the final angles within the specified range ( $[0, 2 π]$ or $[- π, π]$ ).
Verify Solutions: If squaring was used during the process, substitute the values back into the original equation to eliminate extraneous roots.

Question Statement

Background and Explanation

Solution

Q3. Solutions for $ϕ \in [0, 2 π]$

## Part (i)

## Part (ii)

$cos 3 ϕ + cos ϕ = 0$ Using $cos α + cos β = 2 cos (\frac{α + β}{2}) cos (\frac{α - β}{2})$ : $2 cos (2 ϕ) cos (ϕ) = 0$

$cos 2 ϕ = 0 \Rightarrow 2 ϕ = \frac{π}{2}, \frac{3 π}{2} \Rightarrow ϕ = \frac{π}{4}, \frac{3 π}{4}$
$cos ϕ = 0 \Rightarrow ϕ = \frac{π}{2}, \frac{3 π}{2}$ $S . S = {\frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}, \frac{3 π}{2}}$

## Part (iii)

$sin 4 ϕ + sin 3 ϕ = 0$ Using sum-to-product formula: $2 sin (\frac{7 ϕ}{2}) cos (\frac{ϕ}{2}) = 0$

$sin (\frac{7 ϕ}{2}) = 0 \Rightarrow \frac{7 ϕ}{2} = 0, π, 2 π \Rightarrow ϕ = 0, \frac{2 π}{7}, \frac{4 π}{7}$
$cos (\frac{ϕ}{2}) = 0 \Rightarrow \frac{ϕ}{2} = \frac{π}{2} \Rightarrow ϕ = π$ $S . S = {0, π, \frac{2 π}{7}, \frac{4 π}{7}}$

## Part (iv)

$sin 3 ϕ + sin ϕ = sin 2 ϕ$ $2 sin 2 ϕ cos ϕ = sin 2 ϕ$ $sin 2 ϕ (2 cos ϕ - 1) = 0$

$sin 2 ϕ = 0 \Rightarrow 2 ϕ = 0, π, 2 π \Rightarrow ϕ = 0, \frac{π}{2}, π$
$2 cos ϕ = 1 \Rightarrow cos ϕ = \frac{1}{2} \Rightarrow ϕ = \frac{π}{3}, \frac{5 π}{3}$ $S . S = {0, \frac{π}{2}, π, \frac{π}{3}, \frac{5 π}{3}}$

## Part (v)

$4 ϕ = \frac{π}{6} \Rightarrow ϕ = \frac{π}{24}$
$4 ϕ = π - \frac{π}{6} = \frac{5 π}{6} \Rightarrow ϕ = \frac{5 π}{24}$ $S . S = {\frac{π}{24}, \frac{5 π}{24}}$

## Part (vi)

$sin 5 ϕ + sin 3 ϕ + sin 2 ϕ = 0$ $2 sin 4 ϕ cos ϕ + sin 2 ϕ = 0$ $2 (2 sin 2 ϕ cos 2 ϕ) cos ϕ + sin 2 ϕ = 0$ $sin 2 ϕ [4 cos 2 ϕ cos ϕ + 1] = 0$

$sin 2 ϕ = 0 \Rightarrow ϕ = 0, \frac{π}{2}, π$
$4 (2 cos^{2} ϕ - 1) cos ϕ + 1 = 0 \Rightarrow 8 cos^{3} ϕ - 4 cos ϕ + 1 = 0$ By inspection, $cos ϕ = \frac{1}{2}$ is a root. Using synthetic division: Depressed equation: $8 cos^{2} ϕ + 4 cos ϕ - 2 = 0$ .

From $cos ϕ = \frac{1}{2}$ , $ϕ = \frac{π}{3}, \frac{5 π}{3}$ .
From quadratic: $cos ϕ = \frac{- 1 \pm 5}{4}$ .
$cos ϕ = \frac{- 1 + 5}{4} \Rightarrow ϕ = 7 2^{\circ} = \frac{2 π}{5}$ .
$cos ϕ = \frac{- 1 - 5}{4} \Rightarrow ϕ = 14 4^{\circ} = \frac{4 π}{5}$ . $S . S = {0, \frac{π}{2}, π, \frac{π}{3}, \frac{5 π}{3}, \frac{2 π}{5}, \frac{4 π}{5}}$

## Part (vii)

$5 cos ϕ + 4 = 2 (1 - cos^{2} ϕ)$ $2 cos^{2} ϕ + 5 cos ϕ + 2 = 0$ $(2 cos ϕ + 1) (cos ϕ + 2) = 0$

$cos ϕ = - 2$ (Impossible)
$cos ϕ = - 1/2 \Rightarrow ϕ = π - \frac{π}{3}, π + \frac{π}{3} \Rightarrow ϕ = \frac{2 π}{3}, \frac{4 π}{3}$ $S . S = {\frac{2 π}{3}, \frac{4 π}{3}}$

## Part (viii)

$(2 cos^{2} ϕ - 1) + 3 = 5 cos ϕ$ $2 cos^{2} ϕ - 5 cos ϕ + 2 = 0$ $(2 cos ϕ - 1) (cos ϕ - 2) = 0$

$cos ϕ = 2$ (Impossible)
$cos ϕ = 1/2 \Rightarrow ϕ = \frac{π}{3}, \frac{5 π}{3}$ $S . S = {\frac{π}{3}, \frac{5 π}{3}}$

## Part (ix)

## Part (x)

$(cos ϕ - 1)^{2} = 0 \Rightarrow cos ϕ = 1$ $ϕ = 0, 2 π$ $S . S = {0, 2 π}$

## Part (xi)

$tan^{2} ϕ - tan ϕ - 6 = 0$ $(tan ϕ - 3) (tan ϕ + 2) = 0$

$tan ϕ = 3 \Rightarrow ϕ = \frac{71.56 π}{180}, \frac{251.56 π}{180}$
$tan ϕ = - 2 \Rightarrow ϕ = \frac{116.57 π}{180}, \frac{296.57 π}{180}$ $S . S = {\frac{71.56 π}{180}, \frac{116.57 π}{180}, \frac{251.56 π}{180}, \frac{296.57 π}{180}}$

Q4. Solutions for $x \in [- π, π]$

## Part (i)

$3 cos x + sin x = 2 \Rightarrow 3 cos x = 2 - sin x$ Squaring: $9 (1 - sin^{2} x) = 4 + sin^{2} x - 4 sin x \Rightarrow 10 sin^{2} x - 4 sin x - 5 = 0$ $sin x = \frac{2 \pm 3 6}{10}$ .

$sin x \approx 0.9348 \Rightarrow x = 69. 2^{\circ}, 110. 8^{\circ}$ . Checking: $x = 69. 2^{\circ}$ satisfies.
$sin x \approx - 0.5348 \Rightarrow x = - 32.3 3^{\circ}, - 147.6 7^{\circ}$ . Checking: $x = - 32.3 3^{\circ}$ satisfies. $S . S = {\frac{- 32.33 π}{180}, \frac{69.2 π}{180}}$

## Part (ii)

$4 cos x + 3 sin x = 2 \Rightarrow 25 sin^{2} x - 12 sin x - 12 = 0$ $sin x = \frac{6 \pm 4 21}{25}$ .

$sin x \approx 0.9732 \Rightarrow x = 76. 7^{\circ}, 103. 3^{\circ}$ . Checking: $x = 103. 3^{\circ}$ satisfies.
$sin x \approx - 0.4932 \Rightarrow x = - 29.5 5^{\circ}, - 150.4 5^{\circ}$ . Checking: $x = - 29.5 5^{\circ}$ satisfies. $S . S = {\frac{- 29.55 π}{180}, \frac{103.3 π}{180}}$

## Part (iii)

$3 sin x - 2 cos x = 1 \Rightarrow 13 sin^{2} x - 6 sin x - 3 = 0$ $sin x = \frac{3 \pm 4 3}{13}$ .

$sin x \approx 0.76 \Rightarrow x = 49.4 6^{\circ}, 130.5 4^{\circ}$ . Checking: $x = 49.4 6^{\circ}$ satisfies.
$sin x \approx - 0.30 \Rightarrow x = - 17.4 6^{\circ}, - 162.5 4^{\circ}$ . Checking: $x = - 162.5 4^{\circ}$ satisfies. $S . S = {\frac{- 162.54 π}{180}, \frac{49.46 π}{180}}$

## Part (iv)

$3 cos x + sin x = 2 \Rightarrow 4 sin^{2} x - 22 sin x - 1 = 0$ $sin x = \frac{2 \pm 6}{4}$ .

$sin x \approx 0.9659 \Rightarrow x = 7 5^{\circ} (\frac{5 π}{12}), 10 5^{\circ}$ . Checking: $x = \frac{5 π}{12}$ satisfies.
$sin x \approx - 0.2588 \Rightarrow x = - 1 5^{\circ} (- \frac{π}{12}), - 16 5^{\circ}$ . Checking: $x = - \frac{π}{12}$ satisfies. $S . S = {\frac{- π}{12}, \frac{5 π}{12}}$

## Part (v)

$7 cos x - 6 sin x = 4 \Rightarrow 85 sin^{2} x + 48 sin x - 33 = 0$ $sin x = \frac{- 24 \pm 7 69}{85}$ .

$sin x \approx 0.4017 \Rightarrow x = 23.6 8^{\circ}$ . Checking: Satisfied.
$sin x \approx - 0.9664 \Rightarrow x = - 75.1 0^{\circ}, - 104. 9^{\circ}$ . Checking: Neither satisfies. $S . S = {\frac{23.68 π}{180}}$

## Part (vi)

$4 cos x + 2 sin x = 5 \Rightarrow 20 sin^{2} x - 45 sin x - 11 = 0$ $sin x = \frac{5 \pm 2 15}{10}$ .

$sin x \approx 0.9982 \Rightarrow x = 86.5 6^{\circ}$ . Checking: Satisfied.
$sin x \approx - 0.5509 \Rightarrow x = - 33.4 3^{\circ}$ . Checking: Satisfied. $S . S = {\frac{- 33.43 π}{180}, \frac{86.56 π}{180}}$

## Part (vii)

$sec x + 5 tan x + 12 = 0 \Rightarrow 169 sin^{2} x + 10 sin x - 143 = 0$ $sin x = \frac{- 5 \pm 24 42}{169}$ .

$sin x \approx 0.8907 \Rightarrow x = 117.0 4^{\circ}$ . Checking: Satisfied.
$sin x \approx - 0.9499 \Rightarrow x = - 71.7 8^{\circ}$ . Checking: Satisfied. $S . S = {\frac{- 71.78 π}{180}, \frac{117.04 π}{180}}$

## Part (viii)

$2 sin x cos x - cos x = 0 \Rightarrow cos x (2 sin x - 1) = 0$

$cos x = 0 \Rightarrow x = \frac{π}{2}$ (within domain)
$sin x = 1/2 \Rightarrow x = \frac{π}{6}, \frac{5 π}{6}$ $S . S = {\frac{π}{2}, \frac{π}{6}, \frac{5 π}{6}}$

## Part (ix)

$cos^{2} x - sin^{2} x = 1 \Rightarrow cos 2 x = 1$ $2 x = 0 \Rightarrow x = 0, \pm π$ $S . S = {0, \pm π}$

## Part (x)

$sin x (3 sin x - 2) = 0$

$sin x = 0 \Rightarrow x = 0, \pm π$
$sin x = 2/3 \Rightarrow x = 41.8 1^{\circ}, 138.1 9^{\circ}$ $S . S = {0, \pm π, \frac{41.81 π}{180}, \frac{138.19 π}{180}}$

Key Formulas or Methods Used

Pythagorean Identity: $sin^{2} θ + cos^{2} θ = 1$
Sum-to-Product Formulas:
- $cos α + cos β = 2 cos (\frac{α + β}{2}) cos (\frac{α - β}{2})$
- $sin α + sin β = 2 sin (\frac{α + β}{2}) cos (\frac{α - β}{2})$
Double Angle Formulas:
- $sin 2 θ = 2 sin θ cos θ$
- $cos 2 θ = 2 cos^{2} θ - 1 = 1 - 2 sin^{2} θ$
Quadratic Formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$
Synthetic Division: Used for solving higher-degree polynomial equations in terms of trig ratios.

Summary of Steps

Simplify the Equation: Use trigonometric identities to express the equation in terms of a single trigonometric function or a factorable form.
Solve Algebraically: Treat the trigonometric ratio as a variable (e.g., let $u = sin x$ ) and solve the resulting linear or quadratic equation.
Find Reference Angles: Use the inverse trigonometric function to find the base angle in the first quadrant.
Identify Quadrants: Based on the sign of the ratio, determine which quadrants the solutions lie in.
Apply Domain Constraints: Calculate the final angles within the specified range ( $[0, 2 π]$ or $[- π, π]$ ).
Verify Solutions: If squaring was used during the process, substitute the values back into the original equation to eliminate extraneous roots.