# Really Hard Trig Equation Someone help please?

## Homework Statement

Solve the equation:
cos2x - sinx = (1/2) for [0,2pi).

## Homework Equations

sin^2x + cos^2 = 1

## The Attempt at a Solution

cos^2(2x) - sin^2x = 1
1 = 1 Infinite Solutions?

CompuChip
Homework Helper
$$(\cos 2x - \sin x)^2 = \cos^2(2x) - 2 \cos(2x) \sin(x) + \sin^2(x) \neq \cos^2(2x) - \sin^2(x)$$
and
$$\cos^2(2x) - \sin^2(x) \neq \cos^2(x) + \sin^2(x)$$
Maybe
$$\cos(2x) = \cos^2(x) - \sin^2(x)$$
will help?

Please rewrite the question. There is repeated misuse of exponents and misplaced variables. Then I'll help all I can!

cos2x - sinx = (1/2) for [0,2pi).

How does that have misplaced exponents and variables? Would Writing parentheses help? Im really sorry for the confusion and my mistake.

Cos(2x) - sin(x) = (1/2) for [0,2pi).

Cosine of 2x minus sin of x equals 1 half for the range 0 through 2pie

Please rewrite the question. There is repeated misuse of exponents and misplaced variables. Then I'll help all I can!
slap yourself.

Ok, rather than the identity your using ... use:

$$\cos{2x}=1-2\sin^2 x$$

Just factor and it's solved!!!

Last edited:
So then

1 - 2sin^2(x) - sin(x) = (1/2) [-(1/2)]

-2sin^2(x) - sin(x) + (1/2) = 0

I tried it and got sinx = (1 +/- Sqrroot(5)) / -4)
x = -54 and 18.
Somehow that doesn't seem right because radians aren't that big on problems. Any help?

-54 and 18 are the answers in degrees, and they are correct

brendanH the question says between 0 and pi. therefore even if you are right -54 doesnt work

HallsofIvy
And, indeed, since the question itself uses $\pi$, the answer should be in radians, not degrees. There are two solutions between 0 and $\pi$.