Question

How do you find all values of x in the interval [0, 2pi] in the equation `2 sin^2 x = -cos x + 1`?

Answer 1

`x = 0 ,( 2pi)/3 ,( 4pi)/3 , 2pi`

Explanation:

With this type of equation rewrite in terms of cosines

`sin^2 x = ( 1 - cos^2 x )`

the equation is then : `2 (1 - cos^2 x ) = - cosx + 1`

hence `2 - 2cos^2 x = - cosx + 1`

collect all terms to left side and equate to zero.

# 2 - 2 cos^2x + cosx - 1 = 0

ie `2cos^2x - cosx -1 = 0`

factor this quadratic in the same way as algebraic factoring.

Consider factors of -2 that sum to -1 ( middle term).

(They are -2 + 1 = -1)

hence (2cosx + 1 )(cosx - 1 ) = 0

so >2cox + 1 = 0 or cosx - 1 = 0 → `cosx = -1/2 , cosx = 1`

`cosx =1 → x = 0 , 2pi`

`cosx = -1/2 → x =( pi - pi/3) =(2pi)/3 , x =( pi + pi/3) = (4pi)/3`