SOLUTION: Find all values of x in the interval [0, 2π] that satisfy the given equation. -sin2x = (squreroot 2) cos x How do I do this?

Algebra ->  Trigonometry-basics -> SOLUTION: Find all values of x in the interval [0, 2π] that satisfy the given equation. -sin2x = (squreroot 2) cos x How do I do this?      Log On


   

Question 808520: Find all values of x in the interval [0, 2π] that satisfy the given equation.
-sin2x = (squreroot 2) cos x
How do I do this?
Answer by DrBeeee(684) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
(1) -sin2x+=+sqrt%282%29%2Acosx
Use the trig identity for double-angle formulas
(2) sin2x+=+2sinx%2Acosx and get
(3) -2sinx%2Acosx+=+sqrt%282%29%2Acosx
Now as long as cos(x) is not equal to zero (to be verified) we can divide both sides of (3) by cos(x) and get
(4) sinx+=+-1%2Fsqrt%282%29 or
(5) x+=+arcsin%28-1%2Fsqrt%282%29%29 or
(6) x = 225 and 315 degrees.
Since neither angle has it's cosine equal to zero, the division by cosx in (3) was allowed.
Answer: x = {225,315} degrees.