Question

sin3x + cos2x =-2

Answer 1

Solve $sin3x + cos2x = -2$

**Solution:**

As we know $sin\alpha \geq -1$ for any $\alpha$, also $cos\beta \geq -1$ for any $\beta$.

So the maximum value of the expression $sin\alpha + cos\beta \geq -1 - 1$ is $-2$, and it is only if $sin\alpha = -1$ and $cos\beta = -1$

So we have next system:

$$\left\{ \begin{array}{l} sin3x = -1 \\ cos2x = -1 \end{array} \right.$$ $$\left\{ \begin{array}{l} 3x = - \frac {\pi}{2} + 2\pi n, n \in \mathbb {Z} \\ 2x = \pi + 2\pi k, k \in \mathbb {Z} \end{array} \right.$$ $$\left\{ \begin{array}{l} x = - \frac {\pi}{6} + \frac {2\pi n}{3}, n \in \mathbb {Z} \\ x = \frac {\pi}{2} + \pi k, k \in \mathbb {Z} \end{array} \right.$$ $$\left\{ \begin{array}{c} \frac {\pi}{2} + \pi k = - \frac {\pi}{6} + \frac {2\pi n}{3}, n, k \in \mathbb {Z} \\ x = \frac {\pi}{2} + \pi k, k \in \mathbb {Z} \end{array} \right.$$

We must solve the first equation in integers

$$\left\{ \begin{array}{c} k = - \frac {1}{6} + \frac {2n}{3} - \frac {1}{2}, n, k \in \mathbb {Z} \\ x = \frac {\pi}{4} + \pi k, k \in \mathbb {Z} \end{array} \right.$$ $$\left\{ \begin{array}{c} k = \frac {2n - 2}{3}, n, k \in \mathbb {Z} \\ x = \frac {\pi}{2} + \pi k, k \in \mathbb {Z} \end{array} \right.$$ $$\left\{ \begin{array}{c} 3k = 2n - 2, n, k \in \mathbb {Z} \\ x = \frac {\pi}{2} + \pi k, k \in \mathbb {Z} \end{array} \right.$$

The right sight of equation 1 is even, so $3k$ must be even too, so $k$ must be even. If $k$ is even it can be present as $k = 2*p$ where $p \in \mathbb{Z}$

So $x = \frac{\pi}{2} + \pi k = \frac{\pi}{2} + 2\pi p$, $p \in \mathbb{Z}$

**Answer:** $\frac{\pi}{2} + 2\pi p$, $p \in \mathbb{Z}$