Can someone help me with a trig problem?

Question

cos^2x - sin^2x = sinx

i need to solve for x = ~ =...

Answer 1

Remembering the identity, cos²x + sin²x = 1 and rearranging, we see that:

cos²x = 1- sin²x. Let's subsitute this in:

1-sin²x - sin²x = sinx

1-2sin²x = sinx

It may be hard to see, but this is a quadratic. Let y = sinx

1 - 2y² = y

2y² + y - 1 = 0

You can factor this, or use the quadratic eq. to see that

y =1/2, -1

Remember, y = sinx

1/2 = sin(x)---->Take the inverse sin of each side. Ask yourself "what angles gives a sin of 1/2"

sin^-1(1/2) = x

x = 30°(π/6) and 150° (5π/6)

x = sin^-1(-1)

x = 270° (3π/2)

Usually in these kind of problems, they will give a bound, like 0<x<π   or   0<x<2π

If they did not give a bound like that, your two y values are y=1/2 and y = -1, and the angles (x) are

x= 30°, 150°, and 270°

Hope this helps!