Trigonometry, solving for x on an interval

Question

The question is: Solve for x on the interval [-2pi, 4pi]

a.) 1+sin(x) = 2cos^2x

b.) 2cos^2x-1 = 0

My teacher said there should be 9 answers, I am just kind of unsure of where to start

Answer 1

1 + sin(x) = 2cos²(x) [using the Pythagorean identity sin²(x) + cos²(x) = 1 or cos²(x) = 1 - sin²(x) substitute:

1 + sin(x) = 2(1 - sin²(x))

1 + sin(x) = 2 - 2sin²(x)

2sin²(x) + sin(x) - 1 = 0

Let w = sin(x) and substitute:

2w² + w - 1 = 0 and factor to get:

(2w - 1)(w + 1) = 0 solve to get:

w = 1/2 and w = -1 now, back substitute sin(x) = w to get:

sin(x) = 1/2 and sin(x) = -1

sin(x) = 1/2 at x = π/6 and x = 5π/6

sin(x) = -1 at x = 3π/2

So the solutions on [0, 2π) are x = π/6, x = 5π/6, and x = 3π/2

To expand these solutions to the interval [-2π, 4π] consider where the points are on the unit circle and write them accordingly:

The second problem you could just add one to both sides, divide by 2, then take the square root. You'll get 4 answers on [0, 2π) so expanding it to [-2π, 4π] will add an additional 8 answers for a total of 12.

Answer 2

a)1 +sin x = cos^2xiAs 1 - sin^2x =cos^2x, we can write

1 +sin x = 1 - sin^2x

sin x = -sin^2x.

sinx + sin^2x =0

sinx(1+sinx) =0

sinx =0, -1

On [0, 2pi), sinx =0 at 0,pi and sinx=-1at 3pi/2

Thus, on [-2pi.4pi], we have solutions

(-2pi,-pi,-pi/2,0,pi,3pi/2,2pi,3pi,7pi/2,4pi)

b)2cos^2x-1= 0

2cos^2x = 1

cos^2x = 1/2

cosx=±√2/2

On [0,2pi), the solution is (pi/4,3pi/4,5pi/4,7pi/4); thus, on[-2pi,4pi], the solution is

(-7pi/4,-5pi/4,-3pi/4,-pi/4,pi/4,3pi/4,5pi/4,7pi/4,9pi/4,11pi/4,13pi/4,15pi/4)

Answer 3

a.)

1 + sin x = 2 cos² x

1 + sin x = 2(1 - sin² x)

1 + sin x = 2 - 2 sin² x

Can you continue, solve for x, and answer?

b.)

2 cos² x - 1 = 0

2 cos² x = 1

cos² x = 0.5

Can you continue, solve for x, and answer?