Solving Trig Equations — Question 2

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Question 2

Solve the equation cos⁡(2x)=2cos⁡2(x)−1\cos(2x) = 2\cos^2(x) - 1 for all x∈[0,2π]x \in [0, 2\pi], and verify the identity used.

Original worksheet page 1: question and worked solution for 1-4-002
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Question 2 - Solution

Step 1: Recognize the identity.

The right-hand side is a known double-angle identity: cos⁡(2x)=2cos⁡2(x)−1\cos(2x) = 2\cos^2(x) - 1

So this equation is always true: cos⁡(2x)=cos⁡(2x)\cos(2x) = \cos(2x)

Thus, it holds for all xx.

But the problem is to solve: cos⁡(2x)=2cos⁡2(x)−1\cos(2x) = 2\cos^2(x) - 1

Since it holds for all xx, this is an identity. So the equation is true for all x∈[0,2π]x \in [0, 2\pi]

x∈[0,2π]\boxed{x \in [0, 2\pi]}

Verification of identity:

Start with RHS: 2cos⁡2(x)−1=cos⁡(2x)2\cos^2(x) - 1 = \cos(2x)

This is a standard double-angle identity for cosine: cos⁡(2x)=2cos⁡2(x)−1⇒Identity verified.\cos(2x) = 2\cos^2(x) - 1 \Rightarrow \text{Identity verified.}

Original worksheet page 2: question and worked solution for 1-4-002

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