Solving Trig Equations — Question 3

Question 3 - Solution

We begin by using a known identity: $$\sin(2x) = 2\sin(x)\cos(x) \Rightarrow \sin(x)\cos(x) = \frac{1}{2} \sin(2x)$$

Substitute into the equation: $$\frac{1}{2} \sin(2x) = \frac{1}{2} \Rightarrow \sin(2x) = 1$$

Now solve: $$\sin(2x) = 1 \Rightarrow 2x = \frac{\pi}{2} + 2n\pi \quad \text{for integer } n \Rightarrow x = \frac{\pi}{4} + n\pi$$

Find solutions in $[0, 2\pi]$:

$n = 0 \Rightarrow x = \frac{\pi}{4}$ $n = 1 \Rightarrow x = \frac{5\pi}{4}$ $n = 2 \Rightarrow x = \frac{9\pi}{4} > 2\pi$ → discard

$$\boxed{x = \frac{\pi}{4},\ \frac{5\pi}{4}}$$