precalc math problem

Question

Solve the equation for exact solutions in the interval

​[0°​,360°​). Use an algebraic method.

​(cotθ−1​)(2sinθ+1​)=0

Answer 1

(cotx -1)(2sinx+1) = 0

set each factor = 0

cotx = 1

x = 45 or 225 degrees

2sinx = -1

sinx =-1/2

x = 210 or 330 degrees

four solutions:

x = 45, 210, 225, and 330 degrees

Answer 2

So, we will need to solve for x in (cotx -1) and (2sinx+1). Let's first solve for x in (cotx - 1).

cotx - 1 = 0

cotx = 1

x = cot^-1(1) = 45°

However, since sin and cos are both negative in quadrant 3 (180°-270°) and since we know tan is equal to sin/cos, this makes tan positive then in quadrant 3, so we will also have 180° + 45° = 225°.

Now for (2sinx+1),

2sinx+1 = 0

2sinx = -1

sinx = -1/2

x = sin^-1(-1/2) = 360° - 30° = 330°

Again, since sin is also negative in quadrant 3 (180°-270°), we will also have 180° + 30° = 210°

So, there are four exact solutions, which are x = 45°, x = 210°, x = 225°, and x = 330°