Question

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

please show work I am stuck

Answer 1

The solutions are 0° and 3π

Explanation:

On solving the equation given;

`sin((3\pi)/(2)+x )+ sin((3\pi)/(2)+x ) = -2\\2sin((3\pi)/(2)+x ) = -2\\sin((3\pi)/(2)+x ) = -1\\(3\pi)/(2)+x = sin^(-1)-1\\ (3\pi)/(2)+x = -(\pi)/(2) \\x = -(\pi)/(2) -(3\pi)/(2) \\x = (-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°