Answer:
The solutions are 0° and 3π
Step-by-step explanation:
On solving the equation given;
[tex]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\\[/tex]
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°
