how do you solve this trig equation?

Question

gah all this review for finals is killing me. i'm just a little rusty and i can't remember the equation for how to solve this.

cos 2x - sinx = 0

i know you convert the cos 2x to something but i just cant remember what.

Answer 1

cos2x = cos²x - sin²x

so

cos 2x - sinx = 0 becomes

cos²x - sin²x - sinx = 0

Now we wish to express cos²x in terms of sinx, so that the whole equation is in sinx only.
We use the identity
cos²x + sin²x = 1 ---> cos²x = 1 - sin²x

1 - sin²x - sin²x - sinx = 0

Simplify =>

2sin²x + sinx - 1 = 0

This is now a quadratic equation in sinx!
Put sinx = t.
Then the equation becomes:

2t² + t - 1 = 0

Factor (or use the quadratic formula to find solutions):

(2t - 1)(t + 1) = 0

=> t = 1/2 or t = -1

If t = 1/2, then sinx = 1/2 => x = π/6 or 5π/5
If t = -1, then sinx = -1 => x = 3π/2

So the solutions for x are π/6, 5π/5 and 3π/2.

Hope this helps.

Answer 2

U might want to convert them all to sin angles:

cos 2x - sinx= 0: cos^2 x-sin^2 x -sinx=0 : :

1-sin^2 x - sin^2 x - sinx= 0= 1- 2sin^2 x - sinx= 0

Answer 3

cos 2 x = 1 - 2sin^2 x

1 - 2sin^2 x -sin x = 0

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

( 2 sin x -1 ) ( sin x + 1 ) = 0
2sinx - 1= 0 and sin x +1 = 0
sin x = 1/2 sin x = -1
x = 30 , 150 x = 270

then x = 10 , 150 , 270

Answer 4

Very nice answer M. I would concur that your answer is correct.