Solve the equation for x if 0 ≤ x < 2π. Give your answer in radians using exact values only. (Enter your answers as a comma-separated list.) 2 sin2 x − 3 cos x − 3 = 0

Question

Solve the equation for x if 0 ≤ x < 2π. Give your answer in radians using exact values only. (Enter your answers as a comma-separated list.)

2 sin² x − 3 cos x − 3 = 0

Answer 1

Seeing that the sine is squared should be our hint here: we know the Pythagorean identity sin²(x)+cos²(x) = 1.

Rearranging this we see sin²(x) = 1-cos²(x), which can be used to write our original equation solely in terms of cosines, which yields:

2[1-cos²(x)] -3cos(x) -3 = 0

Some simplification:

2- 2cos²(x) - 3 cos(x) - 3 = 0

-2 cos²(x) - 3 cos(x) -1 = 0

2 cos²(x) + 3 cos(x) +1 = 0

Now this doesn't look like we've improved things all that much, but if we allow a quick renaming: w = cos(x)

we have

2w²+3w+1=0, which we can factor

(2w + 1)(w+1) = 0

By the Zero Product property, either

2w + 1 = 0 or w+1 =0

So w = -1/2 or w = -1

Now don't forget where we came from: w = cos(x)

So we need to solve:

cos(x) = -1/2 or cos(x) = -1

Drawing a unit circle, this means we need to find the points where the horizontal value is -1/2 and where it is -1.

For cos(x) = -1/2, drawing a 30-60-90 right triangle may help you find this happens at 120 degrees and 240 degrees (π/3 and 2π/3).

For cos(x) = -1 you'll find this happens once, halfway around the circle, at π.

Answer 2

2 ( 1 - cos^2) - 3cos - 3 = 0

2 - 2cos^2 - 3cos - 3 = 0

-2cos^2 - 3cos - 1 = 0

2cos^2 + 3cos + 1 = 0

2 T^2 - 3T + 1 = 0 where T = cos

( 2T - 1)(T - 1 ) = 0

2T-1 = 0 ---> T = 1/2 ---> cos = 1/2 ---> x=60,300 ---> x = pi/3 or x = 5*pi/3

T-1 = 0 ---> T = 1 ---> cos = 1 ---> x = 0

The solution set is {0,pi/3. 5*pi/3}