Solving Trig Equations — Question 4

Question 4

Solve the equation $$\cos(2x) + \sin(x) = 0$$ for all $x \in [0, 2\pi]$.

Question 4 - Solution

Step 1: Use a double-angle identity for $\cos(2x)$:

We use: $$\cos(2x) = 1 - 2\sin^2(x)$$

Substitute into the equation: $$1 - 2\sin^2(x) + \sin(x) = 0 \Rightarrow -2\sin^2(x) + \sin(x) + 1 = 0$$

Multiply through by $-1$ to simplify: $$2\sin^2(x) - \sin(x) - 1 = 0$$

Step 2: Solve the quadratic equation in $\sin(x)$:

Let $u = \sin(x)$: $$2u^2 - u - 1 = 0 \Rightarrow u = \frac{1 \pm \sqrt{1^2 + 4(2)(1)}}{2(2)} = \frac{1 \pm \sqrt{9}}{4} = \frac{1 \pm 3}{4}$$

So: $$u = 1 \quad \text{or} \quad u = -\frac{1}{2}$$

Now solve $\sin(x) = 1 \Rightarrow x = \frac{\pi}{2}$

And $\sin(x) = -\frac{1}{2} \Rightarrow x = \frac{7\pi}{6},\ \frac{11\pi}{6}$

Final answer: $$\boxed{x = \frac{\pi}{2},\ \frac{7\pi}{6},\ \frac{11\pi}{6}}$$