Solving Trig Equations — Question 2

Question 2

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

Question 2 - Solution

Step 1: Recognize the identity.

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

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

Thus, it holds for all $x$.

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

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

$$\boxed{x \in [0, 2\pi]}$$

Verification of identity:

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

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