Question

Question #a8415

Answer 1

`x=pi/3 or (5pi)/3`

Explanation:

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

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

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

`2-2cos^2x+cosx-2=0`

`cosx=2cos^2x`

`1=2cosx`

`cosx=1/2`

`x=cos^(-1)(1/2)=pi/3 or (2pi-pi/3)=(5pi)/3`

So, `x=pi/3 or (5pi)/3`

Answer 2

`pi/3; pi/2; (3pi)/2; (5pi)/3`

Explanation:

`2sin^2 x + cos x - 2 = 0`
`2 - 2cos^2 x + cos x - 2 = 0`
`- 2cos^2 x + cos x = 0`
`cos x(-2cos x + 1) = 0`.
Either one of the 2 factors should be zero.
a. cos x = 0
Unit circle gives 2 solutions --> `x = pi/2 and x = (3pi)/2`
b. 2cos x = 1 --> `cos x = 1/2`.
Trig table and unit circle give 2 solutions:
`x = pi/3 and x = (5pi)/3`: