Question

Principal solutions of the equation $sin2x + cos2x = 0$, where $\pi < x < 2\pi$ are,

• A. $\frac{7\pi}{8}, \frac{11\pi}{8}$
• B. $\frac{9\pi}{8}, \frac{13\pi}{8}$
• C. $\frac{11\pi}{8}, \frac{15\pi}{8}$
• D. $\frac{15\pi}{8}, \frac{19\pi}{8}$

Answer 1

Answer: (C)

$\frac{11\pi}{8}, \frac{15\pi}{8}$

Answer 2

We start with the equation:

$$\sin2x + \cos2x = 0$$

This can be rearranged as:

$$\sin2x = -\cos2x \implies \tan2x = -1 \quad (\cos2x \neq 0)$$

Thus, the general solution for $2x$ is:

$$2x = -\frac{\pi}{4} + k\pi, \quad k \in \mathbb{Z}$$

Therefore,

$$x = -\frac{\pi}{8} + \frac{k\pi}{2}$$

Now, we need to determine values of $k$ such that:

$$\pi < x < 2\pi$$

For $k=3$:

$$x = -\frac{\pi}{8} + \frac{3\pi}{2} = -\frac{\pi}{8} + \frac{12\pi}{8} = \frac{11\pi}{8}$$

For $k=4$:

$$x = -\frac{\pi}{8} + \frac{4\pi}{2} = -\frac{\pi}{8} + \frac{16\pi}{8} = \frac{15\pi}{8}$$

These values lie in the required interval.