Solving Trig Equations — Question 7

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Question 7

Solve the equation sin⁡(2x)=3cos⁡(x)\sin(2x) = \sqrt{3} \cos(x) for all x∈[0,2π]x \in [0, 2\pi].

Original worksheet page 1: question and worked solution for 1-4-007
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Question 7 - Solution

Use the identity: sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x)

Substitute into the equation: 2sin⁡(x)cos⁡(x)=3cos⁡(x)2\sin(x)\cos(x) = \sqrt{3}\cos(x)

Subtract 3cos⁡(x)\sqrt{3}\cos(x) from both sides: 2sin⁡(x)cos⁡(x)−3cos⁡(x)=0⇒cos⁡(x)(2sin(x)−3)=02\sin(x)\cos(x) - \sqrt{3}\cos(x) = 0 \Rightarrow \cos(x)\left(2\sin(x) - \sqrt{3}\right) = 0

Now solve each factor:

**Case 1:** cos⁡(x)=0⇒x=π2,3π2\cos(x) = 0 \Rightarrow x = \frac{\pi}{2},\ \frac{3\pi}{2}

**Case 2:** 2sin⁡(x)−3=0⇒sin⁡(x)=322\sin(x) - \sqrt{3} = 0 \Rightarrow \sin(x) = \frac{\sqrt{3}}{2}

sin⁡(x)=32⇒x=π3,2π3\sin(x) = \frac{\sqrt{3}}{2} \Rightarrow x = \frac{\pi}{3},\ \frac{2\pi}{3}

Final Solution: x=π3,2π3,π2,3π2\boxed{x = \frac{\pi}{3},\ \frac{2\pi}{3},\ \frac{\pi}{2},\ \frac{3\pi}{2}}

Original worksheet page 2: question and worked solution for 1-4-007

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