In the interval

Question

0 'less than or equal to' x 'less than or equal to' pi
find the values that satisfy the equation
cos 2x - sin x = 0

Answer 1

why not write 0 <= x <= pi

cos2x - sinx = 0
1 - 2sin^2 x - sinx = 0
(2sinx-1)(sinx+1) = 0

not so hard now, eh?

Answer 2

The answer options I get are either
a. 0, pi/3
b. 0, pi/6, pi/3, pi/2

which one would you say suits best?

Answer 3

sinx = 1/2 at x = pi/6, 5pi/6
sinx = -1 at x = 3pi/2, but that's not in the domain.

where did you come up with 0,pi/3,pi/2?

so, pi/6, 5pi/6 are the only solutions. See

http://www.wolframalpha.com/input/?i=cos2x+-+sinx+%3D+0