Question Statement

Answer the following questions using periodicity, even/odd properties, and translation properties of trigonometric functions:

(i) Find the general solution for $\cos x=\frac{1}{\sqrt{2}}$. (ii) Find the general solution for $x$ where the base function involves $\sin x$ such that $x=\frac{3\pi }{2}+2n\pi$, then solve for $\theta$ by replacing $x$ with $2\theta$. (iii) Find $\theta$ in the interval $[0,2\pi )$ such that $\sin \theta =\sin \left(\frac{2\pi }{3}\right)$. (iv) Determine the general solution for $\sin \theta =\sin (3\theta )$. (v) Determine $\theta$ such that $\sin (3\theta )=0$. (vi) Determine the general solution for $\cos (2\theta )=\frac{\sqrt{2}}{2}$. (vii) Find all values of $\theta$ in the interval $[0,2\pi )$ such that $\tan \left(\theta +\frac{\pi }{4}\right)=0$. (viii) Determine the general solution for $\tan (2x)=\sqrt{3}$. (ix) Determine $\theta$ in $[0,2\pi )$ such that $\tan (3\theta )=-\frac{1}{\sqrt{3}}$. (x) Determine the general solution for $\sec (2x)=-2$. (xi) Solve for $\theta$ if $\sin (\theta +\pi )=-\frac{1}{2}$. (xii) Find the values of $\theta$ where $\sin \left(\theta -\frac{\pi }{6}\right)=\sin \left(\theta +\frac{\pi }{6}\right)$. (xiii) Find all values of $\theta$ in the interval $[0,2\pi )$ such that $\cos \left(\theta -\frac{\pi }{3}\right)=\frac{1}{2}$.

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Background and Explanation

To solve trigonometric equations, we identify the reference angle and determine the quadrants where the function has the given sign. We then apply the period of the function ($2\pi$ for $\sin ,\cos ,\sec ,\csc$ and $\pi$ for $\tan ,\cot$) to find the general solution.

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Solution

Part (i)

Equation: $\cos x=\frac{1}{\sqrt{2}}$

As $\cos x$ is positive, the angle lies in the $1^{\text{st}}$ or $4^{\text{th}}$ quadrant. The reference angle is $\frac{\pi }{4}$.

In $1^{\text{st}}$ quadrant: $x=\frac{\pi }{4}$ General value: $x=\frac{\pi }{4}+2m\pi$

In $4^{\text{th}}$ quadrant: $x=2\pi -\frac{\pi }{4}=\frac{8\pi -\pi }{4}=\frac{7\pi }{4}$ General value: $x=\frac{7\pi }{4}+2m\pi$

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Part (ii)

Equation logic: Based on $\sin x$ with period $2\pi$.

$x=\frac{3\pi }{2}+n(2\pi )=\frac{3\pi }{2}+2n\pi$ Replace $x$ by $2\theta$: $2\theta =\frac{3\pi }{2}+2m\pi$ $\theta =\frac{3\pi }{4}+m\pi$ General solution: $\left\{\frac{3\pi }{4}+m\pi \right\}$

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Part (iii)

Equation: $\sin \theta =\sin \left(\frac{2\pi }{3}\right)$

As $\sin \theta$ is positive, $\theta$ lies in the $1^{\text{st}}$ or $2^{\text{nd}}$ quadrant. Reference angle $=\frac{\pi }{3}$.

In $1^{\text{st}}$ quadrant: $\theta =\frac{\pi }{3}$

In $2^{\text{nd}}$ quadrant: $\theta =\pi -\frac{\pi }{3}=\frac{2\pi }{3}$ Solution set in $[0,2\pi )$: $\left\{\frac{\pi }{3},\frac{2\pi }{3}\right\}$

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Part (iv)

Equation: $\sin \theta =\sin 3\theta$

$\sin \theta -\sin 3\theta =0$ Using sum-to-product formula: $2\cos \left(\frac{\theta +3\theta }{2}\right)\sin \left(\frac{\theta -3\theta }{2}\right)=0$ $2\cos \left(\frac{4\theta }{2}\right)\sin \left(\frac{-2\theta }{2}\right)=0$ $2\cos (2\theta )\cdot \sin (-\theta )=0$ Since $\sin (-\theta )=-\sin \theta$ (odd function): $-2\cos 2\theta \sin \theta =0$

Case 1: $\sin \theta =0$ $\theta ={\sin }^{-1}(0)⟹\theta =n\pi$

Case 2: $\cos 2\theta =0$ $2\theta ={\cos }^{-1}(0)⟹2\theta =(2n+1)\frac{\pi }{2}$ $\theta =(2n+1)\frac{\pi }{4}$ General solution: $\{n\pi \}\cup \left\{(2n+1)\frac{\pi }{4}\right\}$

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Part (v)

Equation: $\sin 3\theta =0$

Let $x=3\theta$. The base function is $\sin x=0$. $x={\sin }^{-1}(0)⟹x=n\pi$ Replace $x$ by $3\theta$: $3\theta =n\pi ⟹\theta =\frac{n\pi }{3}$

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Part (vi)

Equation: $\cos 2\theta =\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

Let $x=2\theta$. Base function: $\cos x=\frac{1}{\sqrt{2}}$. $\cos x$ is positive, so $x$ is in $1^{\text{st}}$ or $4^{\text{th}}$ quadrant. Reference angle $=\frac{\pi }{4}$.

In $1^{\text{st}}$ quadrant: $x=\frac{\pi }{4}+2n\pi$ Replace $x$ by $2\theta$: $2\theta =\frac{\pi }{4}+2m\pi ⟹\theta =\frac{\pi }{8}+m\pi$

In $4^{\text{th}}$ quadrant: $x=2\pi -\frac{\pi }{4}=\frac{7\pi }{4}$ General value: $x=\frac{7\pi }{4}+2n\pi$ Replace $x$ by $2\theta$: $2\theta =\frac{7\pi }{4}+2n\pi ⟹\theta =\frac{7\pi }{8}+n\pi$ General solution: $\left\{\frac{\pi }{8}+m\pi \right\}\cup \left\{\frac{7\pi }{8}+n\pi \right\}$

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Part (vii)

Equation: $\tan \left(\theta +\frac{\pi }{4}\right)=0$ in $[0,2\pi )$

Let $x=\theta +\frac{\pi }{4}$. Base function: $\tan x=0$. $x=0,\pi ,2\pi$

If $x=\pi$: $\theta +\frac{\pi }{4}=\pi ⟹\theta =\pi -\frac{\pi }{4}=\frac{3\pi }{4}$

If $x=2\pi$: $\theta +\frac{\pi }{4}=2\pi ⟹\theta =2\pi -\frac{\pi }{4}=\frac{7\pi }{4}$ Solution Set: $\left\{\frac{3\pi }{4},\frac{7\pi }{4}\right\}$

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Part (viii)

Equation: $\tan (2x)=\sqrt{3}$

Let $\theta =2x$. $\tan \theta$ is positive, so $\theta$ is in $1^{\text{st}}$ or $3^{\text{rd}}$ quadrant. Reference angle $=\frac{\pi }{3}$.

In $1^{\text{st}}$ quadrant: $\theta =\frac{\pi }{3}+n\pi$ Replace $\theta$ by $2x$: $2x=\frac{\pi }{3}+n\pi ⟹x=\frac{\pi }{6}+\frac{n\pi }{2}$

In $3^{\text{rd}}$ quadrant: $\theta =\pi +\frac{\pi }{3}=\frac{4\pi }{3}$ General value: $\theta =\frac{4\pi }{3}+n\pi$ Replace $\theta$ by $2x$: $2x=\frac{4\pi }{3}+n\pi ⟹x=\frac{2\pi }{3}+\frac{n\pi }{2}$ General solution: $\left\{\frac{\pi }{6}+\frac{n\pi }{2}\right\}\cup \left\{\frac{2\pi }{3}+\frac{n\pi }{2}\right\}$

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Part (ix)

Equation: $\tan (3\theta )=-\frac{1}{\sqrt{3}}$ in $[0,2\pi )$

Let $x=3\theta$. $\tan x$ is negative, so $x$ is in $2^{\text{nd}}$ or $4^{\text{th}}$ quadrant. Reference angle $=\frac{\pi }{6}$.

In $2^{\text{nd}}$ quadrant: $x=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$ Replace $x$ by $3\theta$: $3\theta =\frac{5\pi }{6}⟹\theta =\frac{5\pi }{18}$

In $4^{\text{th}}$ quadrant: $x=2\pi -\frac{\pi }{6}=\frac{11\pi }{6}$ Replace $x$ by $3\theta$: $3\theta =\frac{11\pi }{6}⟹\theta =\frac{11\pi }{18}$ Solution Set: $\left\{\frac{5\pi }{18},\frac{11\pi }{18}\right\}$

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Part (x)

Equation: $\sec (2x)=-2$

Convert to cosine: $\cos (2x)=-\frac{1}{2}$. Let $\theta =2x$. $\cos \theta$ is negative, so $\theta$ is in $2^{\text{nd}}$ or $3^{\text{rd}}$ quadrant. Reference angle $=\frac{\pi }{3}$.

In $2^{\text{nd}}$ quadrant: $\theta =\pi -\frac{\pi }{3}=\frac{2\pi }{3}$ General value: $\theta =\frac{2\pi }{3}+2n\pi$ Replace $\theta$ by $2x$: $2x=\frac{2\pi }{3}+2n\pi ⟹x=\frac{\pi }{3}+n\pi$

In $3^{\text{rd}}$ quadrant: $\theta =\pi +\frac{\pi }{3}=\frac{4\pi }{3}$ General value: $\theta =\frac{4\pi }{3}+2n\pi$ Replace $\theta$ by $2x$: $2x=\frac{4\pi }{3}+2n\pi ⟹x=\frac{2\pi }{3}+n\pi$ General solution: $\left\{\frac{\pi }{3}+n\pi \right\}\cup \left\{\frac{2\pi }{3}+n\pi \right\}$

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Part (xi)

Equation: $\sin (\theta +\pi )=-\frac{1}{2}$

Let $x=\theta +\pi$. $\sin x=-\frac{1}{2}$. $\sin x$ is negative, so $x$ is in $3^{\text{rd}}$ or $4^{\text{th}}$ quadrant. Reference angle $=\frac{\pi }{6}$.

In $3^{\text{rd}}$ quadrant: $x=\pi +\frac{\pi }{6}=\frac{7\pi }{6}$ Replace $x$ by $\theta +\pi$: $\theta +\pi =\frac{7\pi }{6}⟹\theta =\frac{7\pi }{6}-\pi =\frac{\pi }{6}$

In $4^{\text{th}}$ quadrant: $x=2\pi -\frac{\pi }{6}=\frac{11\pi }{6}$ Replace $x$ by $\theta +\pi$: $\theta +\pi =\frac{11\pi }{6}⟹\theta =\frac{11\pi }{6}-\pi =\frac{5\pi }{6}$ Solution Set: $\left\{\frac{\pi }{6},\frac{5\pi }{6}\right\}$

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Part (xii)

Equation: $\sin \left(\theta -\frac{\pi }{6}\right)=\sin \left(\theta +\frac{\pi }{6}\right)$

$\sin \left(\theta -\frac{\pi }{6}\right)-\sin \left(\theta +\frac{\pi }{6}\right)=0$ Using sum-to-product: $2\cos \left(\frac{\theta -\frac{\pi }{6}+\theta +\frac{\pi }{6}}{2}\right)\sin \left(\frac{\theta -\frac{\pi }{6}-(\theta +\frac{\pi }{6})}{2}\right)=0$ $2\cos \left(\frac{2\theta }{2}\right)\sin \left(\frac{-2\pi /6}{2}\right)=0$ $2\cos \theta \sin \left(-\frac{\pi }{6}\right)=0$ $2\cos \theta \left(-\frac{1}{2}\right)=0$ $-\cos \theta =0⟹\cos \theta =0$ $\theta =(2n+1)\frac{\pi }{2}\text{ or }\theta =n\pi +\frac{\pi }{2}$

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Part (xiii)

Equation: $\cos \left(\theta -\frac{\pi }{3}\right)=\frac{1}{2}$ in $[0,2\pi )$

Let $x=\theta -\frac{\pi }{3}$. $\cos x=\frac{1}{2}$. $\cos x$ is positive, so $x$ is in $1^{\text{st}}$ or $4^{\text{th}}$ quadrant. Reference angle $=\frac{\pi }{3}$.

In $1^{\text{st}}$ quadrant: $x=\frac{\pi }{3}$ Replace $x$ by $\theta -\frac{\pi }{3}$: $\theta -\frac{\pi }{3}=\frac{\pi }{3}⟹\theta =\frac{2\pi }{3}$

In $4^{\text{th}}$ quadrant: $x=2\pi -\frac{\pi }{3}=\frac{5\pi }{3}$ Replace $x$ by $\theta -\frac{\pi }{3}$: $\theta -\frac{\pi }{3}=\frac{5\pi }{3}⟹\theta =\frac{5\pi }{3}+\frac{\pi }{3}=2\pi$ Since the interval is $[0,2\pi )$, $2\pi$ is excluded. Solution Set: $\left\{\frac{2\pi }{3}\right\}$

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Key Formulas or Methods Used

• Reference Angle: The acute angle formed with the x-axis.
• General Solution for Sine/Cosine: $\theta +2n\pi$.
• General Solution for Tangent: $\theta +n\pi$.
• Sum-to-Product Formula: $\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$.
• Odd Function Property: $\sin (-\theta )=-\sin \theta$.

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Summary of Steps

1. Identify the trigonometric function and its value.
2. Determine the reference angle and the quadrants based on the sign (+/-).
3. Find the specific angles within one period.
4. Add the appropriate multiple of the period ($2n\pi$ or $n\pi$) for general solutions.
5. If the argument is modified (e.g., $2\theta$ or $\theta +\pi$), solve for $\theta$ algebraically.
6. Filter results if a specific interval like $[0,2\pi )$ is required.