Question

Find the solutions of each equation on the interval [0, 2π). si (3pi/2+x)+ sin (3pi/2+x)=-2

Answer 1

Answer:

The solutions are 0° and 3π

Step-by-step explanation:

On solving the equation given;

$sin(\frac{3\pi}{2}+x )+ sin(\frac{3\pi}{2}+x ) = -2\\2sin(\frac{3\pi}{2}+x ) = -2\\sin(\frac{3\pi}{2}+x ) = -1\\\frac{3\pi}{2}+x = sin^{-1}-1\\ \frac{3\pi}{2}+x = -\frac{\pi}{2} \\x = -\frac{\pi}{2} -\frac{3\pi}{2} \\x = \frac{-4\pi}{2} \\x = -2\pi$

Since sin is negative in the 3rd and 4th quadrant,

In the 3rd quadrant;

x =180°+2π

x = π + 2π

x = 3π

In the 4th quadrant;

x = 360°-2π

x = 2π-2π

x = 0°