Question Statement

Determine the solution for each of the following trigonometric equations over the interval $[0,2\pi )$:

(i) $\sin 3\theta +\sin \theta =0$ (ii) $\sin 2\theta =2\cos \theta$ (iii) $\sin 2\theta +\sin \theta =0$ (iv) $\cos 2\theta +\cos \theta =0$ (v) $2{\sin }^3\theta ={\sin }^2\theta$ (vi) $\cos 2\theta =3\cos \theta -2$ (vii) $\cos 3\theta =2-\cos \theta$ (viii) $\cos 3\theta =\cos 5\theta$ (ix) $\cos \theta \sin \theta =\frac{1}{2}$ (x) ${\cos }^2\theta +4\sin \theta =4$ (xi) $\cos \rho -\sqrt{3}\sin \rho =2$ (xii) $3{\tan }^2\theta -1=0$

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Background and Explanation

To solve trigonometric equations, we use fundamental identities (like double-angle or sum-to-product formulas) to simplify the expression into a single trigonometric ratio. Once simplified, we determine the reference angle and identify the specific quadrants where the solution exists within the given interval $[0,2\pi )$.

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Solution

Part (i)

Given: $\sin 3\theta +\sin \theta =0$

Using the sum-to-product formula: $\sin \alpha +\sin \beta =2\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)$

$\begin{array}{rl}2\sin \left(\frac{3\theta +\theta }{2}\right)\cos \left(\frac{3\theta -\theta }{2}\right)& =0\\ 2\sin \left(\frac{4\theta }{2}\right)\cos \left(\frac{2\theta }{2}\right)& =0\\ 2\sin 2\theta \cos \theta & =0\end{array}$

This implies $\sin 2\theta \cdot \cos \theta =0$. We solve the two cases:

1. $\sin 2\theta =0$ $\begin{array}{rl}2\theta & ={\sin }^{-1}(0)\\ 2\theta & =0\text{ or }2\theta =\pi \\ \theta & =0,\theta =\frac{\pi }{2}\end{array}$

2. $\cos \theta =0$ $\begin{array}{rl}\theta & ={\cos }^{-1}(0)\\ \theta & =\frac{\pi }{2}\text{ or }\theta =\frac{3\pi }{2}\end{array}$

$\therefore \text{Solution Set (S.S)}=\left\{0,\frac{\pi }{2},\frac{3\pi }{2}\right\}$

Part (ii)

Given: $\sin 2\theta =2\cos \theta$

$\begin{array}{rl}\sin 2\theta -2\cos \theta & =0\\ 2\sin \theta \cos \theta -2\cos \theta & =0\\ 2\cos \theta (\sin \theta -1)& =0\end{array}$

Setting each factor to zero:

• $2\cos \theta =0\Rightarrow \cos \theta =0\Rightarrow \theta =\frac{\pi }{2},\frac{3\pi }{2}$
• $\sin \theta -1=0\Rightarrow \sin \theta =1\Rightarrow \theta =\frac{\pi }{2}$

$\therefore \text{S.S}=\left\{\frac{\pi }{2},\frac{3\pi }{2}\right\}$

Part (iii)

Given: $\sin 2\theta +\sin \theta =0$

$\begin{array}{rl}2\sin \theta \cos \theta +\sin \theta & =0\\ \sin \theta (2\cos \theta +1)& =0\end{array}$

1. $\sin \theta =0\Rightarrow \theta =0,\pi$
2. $2\cos \theta +1=0\Rightarrow \cos \theta =-\frac{1}{2}$

Since $\cos \theta$ is negative, $\theta$ lies in the 2nd or 3rd quadrant. The reference angle is $\frac{\pi }{3}$.

• In 2nd quadrant: $\theta =\pi -\frac{\pi }{3}=\frac{2\pi }{3}$
• In 3rd quadrant: $\theta =\pi +\frac{\pi }{3}=\frac{4\pi }{3}$

$\therefore \text{S.S}=\left\{0,\pi ,\frac{2\pi }{3},\frac{4\pi }{3}\right\}$

Part (iv)

Given: $\cos 2\theta +\cos \theta =0$

Using the formula: $\cos \alpha +\cos \beta =2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)$

$\begin{array}{rl}2\cos \left(\frac{2\theta +\theta }{2}\right)\cos \left(\frac{2\theta -\theta }{2}\right)& =0\\ 2\cos \left(\frac{3\theta }{2}\right)\cdot \cos \left(\frac{\theta }{2}\right)& =0\end{array}$

1. $\cos \frac{3\theta }{2}=0\Rightarrow \frac{3\theta }{2}=\frac{\pi }{2},\frac{3\pi }{2}\Rightarrow 3\theta =\pi ,3\pi \Rightarrow \theta =\frac{\pi }{3},\pi$
2. $\cos \frac{\theta }{2}=0\Rightarrow \frac{\theta }{2}=\frac{\pi }{2},\frac{3\pi }{2}\Rightarrow \theta =\pi ,3\pi$

$\therefore \text{S.S}=\left\{\frac{\pi }{3},\pi ,3\pi \right\}$

Part (v)

Given: $2{\sin }^3\theta ={\sin }^2\theta$

$\begin{array}{rl}2{\sin }^3\theta -{\sin }^2\theta & =0\\ {\sin }^2\theta (2\sin \theta -1)& =0\end{array}$

1. ${\sin }^2\theta =0\Rightarrow \sin \theta =0\Rightarrow \theta =0,\pi$
2. $2\sin \theta -1=0\Rightarrow \sin \theta =\frac{1}{2}$

Since $\sin \theta$ is positive, $\theta$ lies in the 1st or 2nd quadrant. Reference angle is $\frac{\pi }{6}$.

• In 1st quadrant: $\theta =\frac{\pi }{6}$
• In 2nd quadrant: $\theta =\pi -\frac{\pi }{6}=\frac{5\pi }{6}$

$\therefore \text{S.S}=\left\{0,\pi ,\frac{\pi }{6},\frac{5\pi }{6}\right\}$

Part (vi)

Given: $\cos 2\theta =3\cos \theta -2$

Using $\cos 2\theta =2{\cos }^2\theta -1$:

$\begin{array}{rl}2{\cos }^2\theta -1& =3\cos \theta -2\\ 2{\cos }^2\theta -3\cos \theta +1& =0\\ 2{\cos }^2\theta -2\cos \theta -\cos \theta +1& =0\\ 2\cos \theta (\cos \theta -1)-1(\cos \theta -1)& =0\\ (\cos \theta -1)(2\cos \theta -1)& =0\end{array}$

1. $\cos \theta -1=0\Rightarrow \cos \theta =1\Rightarrow \theta =0$ (Note: $2\pi$ is excluded from $[0,2\pi )$)
2. $2\cos \theta -1=0\Rightarrow \cos \theta =\frac{1}{2}$

Since $\cos \theta$ is positive, $\theta$ lies in the 1st or 4th quadrant. Reference angle is $\frac{\pi }{3}$.

• In 1st quadrant: $\theta =\frac{\pi }{3}$
• In 4th quadrant: $\theta =2\pi -\frac{\pi }{3}=\frac{5\pi }{3}$

$\therefore \text{S.S}=\left\{0,\frac{\pi }{3},\frac{5\pi }{3}\right\}$

Part (vii)

Given: $\cos 3\theta =2-\cos \theta$

Using $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$:

$\begin{array}{rl}4{\cos }^3\theta -3\cos \theta & =2-\cos \theta \\ 4{\cos }^3\theta -2\cos \theta -2& =0\dots (1)\end{array}$

Testing $\cos \theta =1$: $4(1{)}^3-2(1)-2=0$, which is true. Using synthetic division for $\cos \theta =1$:

The depressed equation is $4{\cos }^2\theta +4\cos \theta +2=0$. Using the quadratic formula: $\cos \theta =\frac{-4\pm \sqrt{16-32}}{8}=\frac{-4\pm 4i}{8}$ These roots are imaginary and are neglected. Thus, $\cos \theta =1\Rightarrow \theta =0$ ($2\pi$ is excluded).

$\therefore \text{S.S}=\{0\}$

Part (viii)

Given: $\cos 3\theta =\cos 5\theta \Rightarrow \cos 5\theta -\cos 3\theta =0$

Using $\cos \alpha -\cos \beta =-2\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)$:

$\begin{array}{rl}-2\sin \left(\frac{5\theta +3\theta }{2}\right)\sin \left(\frac{5\theta -3\theta }{2}\right)& =0\\ -2\sin 4\theta \sin \theta & =0\end{array}$

1. $\sin 4\theta =0\Rightarrow 4\theta =0,\pi \Rightarrow \theta =0,\frac{\pi }{4}$
2. $\sin \theta =0\Rightarrow \theta =0,\pi$

$\therefore \text{S.S}=\left\{0,\frac{\pi }{4},\pi \right\}$

Part (ix)

Given: $\cos \theta \sin \theta =\frac{1}{2}$

$\begin{array}{rl}2\cos \theta \sin \theta & =1\\ \sin 2\theta & =1\\ 2\theta & =\frac{\pi }{2}\\ \theta & =\frac{\pi }{4}\end{array}$

$\therefore \text{S.S}=\left\{\frac{\pi }{4}\right\}$

Part (x)

Given: ${\cos }^2\theta +4\sin \theta =4$

$\begin{array}{rl}(1-{\sin }^2\theta )+4\sin \theta -4& =0\\ -{\sin }^2\theta +4\sin \theta -3& =0\\ {\sin }^2\theta -4\sin \theta +3& =0\\ (\sin \theta -3)(\sin \theta -1)& =0\end{array}$

1. $\sin \theta =3$ (Impossible, as $-1\le \sin \theta \le 1$)
2. $\sin \theta =1\Rightarrow \theta =\frac{\pi }{2}$

$\therefore \text{S.S}=\left\{\frac{\pi }{2}\right\}$

Part (xi)

Given: $\cos \rho -\sqrt{3}\sin \rho =2$

$\cos \rho -2=\sqrt{3}\sin \rho$

Squaring both sides: $\begin{array}{rl}(\cos \rho -2{)}^2& =3{\sin }^2\rho \\ {\cos }^2\rho -4\cos \rho +4& =3(1-{\cos }^2\rho )\\ 4{\cos }^2\rho -4\cos \rho +1& =0\\ (2\cos \rho -1{)}^2& =0\\ \cos \rho & =\frac{1}{2}\end{array}$

Reference angle is $\frac{\pi }{3}$.

• In 1st quadrant: $\rho =\frac{\pi }{3}$
• In 4th quadrant: $\rho =2\pi -\frac{\pi }{3}=\frac{5\pi }{3}$

$\therefore \text{S.S}=\left\{\frac{\pi }{3},\frac{5\pi }{3}\right\}$

Part (xii)

Given: $3{\tan }^2\theta -1=0$

$\begin{array}{r}{\tan }^2\theta =\frac{1}{3}\Rightarrow \tan \theta =\pm \frac{1}{\sqrt{3}}\end{array}$

Reference angle is $\frac{\pi }{6}$.

• For $\tan \theta =\frac{1}{\sqrt{3}}$ (1st and 3rd quadrants): $\theta =\frac{\pi }{6},\frac{7\pi }{6}$
• For $\tan \theta =-\frac{1}{\sqrt{3}}$ (2nd and 4th quadrants): $\theta =\frac{5\pi }{6},\frac{11\pi }{6}$

$\therefore \text{S.S}=\left\{\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}\right\}$

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Key Formulas or Methods Used

• Sum-to-Product Formulas:
  • $\sin \alpha +\sin \beta =2\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)$
  • $\cos \alpha +\cos \beta =2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)$
  • $\cos \alpha -\cos \beta =-2\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)$
• Double Angle Formulas: $\sin 2\theta =2\sin \theta \cos \theta$, $\cos 2\theta =2{\cos }^2\theta -1$
• Triple Angle Formula: $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$
• Pythagorean Identity: ${\cos }^2\theta +{\sin }^2\theta =1$
• Synthetic Division: Used to factor higher-degree polynomials.

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Summary of Steps

1. Simplify the equation using trigonometric identities to isolate a single function or factorable expression.
2. Set each factor to zero to find basic trigonometric equations (e.g., $\sin \theta =k$).
3. Find the reference angle using the inverse trigonometric function.
4. Determine the specific solutions within the interval $[0,2\pi )$ by applying the reference angle to the appropriate quadrants based on the sign of the function.
5. Check for extraneous solutions (especially when squaring both sides) and ensure all values fall within the required interval.