Question Statement

Solve the following trigonometric equations within the domain $[0,2\pi ]$:

(i) $\sqrt{2}\sin x=1$ (ii) $5\sin x=2$ (iii) $2\cos \theta +1=0$ (iv) $\cot x+\sqrt{3}=0$ (v) $\cos x=0.71$ (vi) $\sqrt{6}\csc x=\sqrt{8}$ (vii) $4\cos x-3=0$ (viii) $\sqrt{2}\csc x+2=0$ (ix) ${\tan }^2\rho =3$

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Background and Explanation

To solve these equations, we first isolate the trigonometric function. We then determine the reference angle (the acute angle relative to the x-axis) and identify which quadrants the solutions lie in based on whether the value is positive or negative (using the CAST rule). Finally, we calculate the specific angles within the $[0,2\pi ]$ range.

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Solution

Part (i)

$\begin{array}{rl}\sqrt{2}\sin x& =1\\ \sin x& =\frac{1}{\sqrt{2}}\end{array}$ Since $\sin x$ is positive, the angle lies in the $1^{\text{st}}$ or $2^{\text{nd}}$ quadrant. The reference angle is $\frac{\pi }{4}$.

• In 1st quadrant: $x=\frac{\pi }{4}$
• In 2nd quadrant: $x=\pi -\frac{\pi }{4}=\frac{3\pi }{4}$

Solution Set: $S.S=\left\{\frac{\pi }{4},\frac{3\pi }{4}\right\}$

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Part (ii)

$\begin{array}{rl}5\sin x& =2\\ \sin x& =\frac{2}{5}\end{array}$ Since $\sin x$ is positive, the angle lies in the $1^{\text{st}}$ or $2^{\text{nd}}$ quadrant. The reference angle is $x={\sin }^{-1}\left(\frac{2}{5}\right)=23.57^{\circ }$. Converting to radians: $23.57\times \frac{\pi }{180}=\frac{23.57\pi }{180}\text{ rad}$.

• In 1st quadrant: $x=\frac{23.57\pi }{180}$
• In 2nd quadrant: $x=\pi -\frac{23.57\pi }{180}=\frac{156.43\pi }{180}$

Solution Set: $S.S=\left\{\frac{23.57\pi }{180},\frac{156.43\pi }{180}\right\}$

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Part (iii)

$\begin{array}{rl}2\cos \theta +1& =0\\ 2\cos \theta & =-1\\ \cos \theta & =-\frac{1}{2}\end{array}$ Since $\cos \theta$ is negative, $\theta$ lies in the $2^{\text{nd}}$ or $3^{\text{rd}}$ quadrant. The reference angle is $\frac{\pi }{3}$.

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

Solution Set: $S.S=\left\{\frac{2\pi }{3},\frac{4\pi }{3}\right\}$

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Part (iv)

$\begin{array}{rl}\cot x+\sqrt{3}& =0\\ \cot x& =-\sqrt{3}\\ \tan x& =-\frac{1}{\sqrt{3}}\end{array}$ Since $\tan x$ is negative, the angle lies in the $2^{\text{nd}}$ or $4^{\text{th}}$ quadrant. The reference angle is $\frac{\pi }{6}$.

• In 2nd quadrant: $x=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$
• In 4th quadrant: $x=2\pi -\frac{\pi }{6}=\frac{11\pi }{6}$

Solution Set: $S.S=\left\{\frac{5\pi }{6},\frac{11\pi }{6}\right\}$

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Part (v)

$\cos x=0.71$ Since $\cos x$ is positive, the angle lies in the $1^{\text{st}}$ or $4^{\text{th}}$ quadrant. The reference angle is $x={\cos }^{-1}(0.71)\approx 44.765^{\circ }\approx 45^{\circ }$. In radians, $45^{\circ }=\frac{\pi }{4}\text{ rad}$.

• In 1st quadrant: $x=\frac{\pi }{4}$
• In 4th quadrant: $x=2\pi -\frac{\pi }{4}=\frac{7\pi }{4}$

Solution Set: $S.S=\left\{\frac{\pi }{4},\frac{7\pi }{4}\right\}$

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Part (vi)

$\begin{array}{rl}\sqrt{6}\csc x& =\sqrt{8}\\ \csc x& =\frac{\sqrt{8}}{\sqrt{6}}=\sqrt{\frac{8}{6}}=\sqrt{\frac{4}{3}}\\ \csc x& =\frac{2}{\sqrt{3}}\\ \sin x& =\frac{\sqrt{3}}{2}\end{array}$ Since $\sin x$ is positive, the angle lies in the $1^{\text{st}}$ or $2^{\text{nd}}$ quadrant. The reference angle is $\frac{\pi }{3}$.

• In 1st quadrant: $x=\frac{\pi }{3}$
• In 2nd quadrant: $x=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$

Solution Set: $S.S=\left\{\frac{\pi }{3},\frac{2\pi }{3}\right\}$

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Part (vii)

$\begin{array}{rl}4\cos x-3& =0\\ 4\cos x& =3\\ \cos x& =\frac{3}{4}\end{array}$ Since $\cos x$ is positive, the angle lies in the $1^{\text{st}}$ or $4^{\text{th}}$ quadrant. The reference angle is $x={\cos }^{-1}\left(\frac{3}{4}\right)=41.41^{\circ }$. In radians: $\frac{41.41\pi }{180}\text{ rad}$.

• In 1st quadrant: $x=\frac{41.41\pi }{180}$
• In 4th quadrant: $x=2\pi -\frac{41.41\pi }{180}=\frac{318.59\pi }{180}$

Solution Set: $S.S=\left\{\frac{41.41\pi }{180},\frac{318.59\pi }{180}\right\}$

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Part (viii)

$\begin{array}{rl}\sqrt{2}\csc x+2& =0\\ \sqrt{2}\csc x& =-2\\ \csc x& =-\frac{2}{\sqrt{2}}=-\sqrt{2}\\ \sin x& =-\frac{1}{\sqrt{2}}\end{array}$ Since $\sin x$ is negative, the angle lies in the $3^{\text{rd}}$ or $4^{\text{th}}$ quadrant. The reference angle is $\frac{\pi }{4}$.

• In 3rd quadrant: $x=\pi +\frac{\pi }{4}=\frac{5\pi }{4}$
• In 4th quadrant: $x=2\pi -\frac{\pi }{4}=\frac{7\pi }{4}$

Solution Set: $S.S=\left\{\frac{5\pi }{4},\frac{7\pi }{4}\right\}$

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Part (ix)

${\tan }^2\rho =3$ Taking the square root of both sides: $\tan \rho =\pm \sqrt{3}$

Case 1: $\tan \rho =\sqrt{3}$ (Positive) Angle lies in $1^{\text{st}}$ or $3^{\text{rd}}$ quadrant. Reference angle is $\frac{\pi }{3}$.

• 1st quad: $\rho =\frac{\pi }{3}$
• 3rd quad: $\rho =\pi +\frac{\pi }{3}=\frac{4\pi }{3}$

Case 2: $\tan \rho =-\sqrt{3}$ (Negative) Angle lies in $2^{\text{nd}}$ or $4^{\text{th}}$ quadrant. Reference angle is $\frac{\pi }{3}$.

• 2nd quad: $\rho =\pi -\frac{\pi }{3}=\frac{2\pi }{3}$
• 4th quad: $\rho =2\pi -\frac{\pi }{3}=\frac{5\pi }{3}$

Solution Set: $S.S=\left\{\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}\right\}$

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Key Formulas or Methods Used

• Reciprocal Identities: $\sin x=\frac{1}{\csc x}$, $\cos x=\frac{1}{\sec x}$, $\tan x=\frac{1}{\cot x}$.
• Reference Angle (${\theta }_{ref}$): The acute angle formed with the x-axis.
• Quadrant Rules for $[0,2\pi ]$:
  • Quadrant I: $\theta ={\theta }_{ref}$
  • Quadrant II: $\theta =\pi -{\theta }_{ref}$
  • Quadrant III: $\theta =\pi +{\theta }_{ref}$
  • Quadrant IV: $\theta =2\pi -{\theta }_{ref}$
• Degree to Radian Conversion: $\text{Radians}=\text{Degrees}\times \frac{\pi }{180}$.

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Summary of Steps

1. Isolate the trigonometric function (e.g., $\sin x=a$).
2. Determine the sign of the value to identify the possible quadrants.
3. Find the reference angle using the inverse function of the absolute value.
4. Apply the quadrant formulas to find all angles within $[0,2\pi ]$.
5. Combine all valid angles into the Solution Set ($S.S$).