Find all real numbers x such that 2sin^2 x+7cosx+2=0?

Answer 1

2sin^2 x+7cosx+2=0
2(1-cos^2x) +7cosx +2 = 0
-2cos^2 x +7cosx +4 = 0
cosx = [-7 +/- sqrt(49 -4(-2)(4))]/(2(-2))
cosx = [-7 +/- 9]/-4
cos x = -1/2 or 4 (4 is rejected as extraneous)
So x = 120 degrees or 240 degrees
or x = 2pi/3 radians or 4pi/3 radians
You can add 2npi to either of the last two answers to get additional answers. n = any positive or negative integer.

Answer 2

2(1-cos^2x)+7cosx+2=0
-2cos^2x+7cosx+4=0
2cos^2x-7cosx-4=0
(2cosx+1)(cosx-4)=0
cosx=-.5 and cosx can never equal 4.
x= 2pi/3 +2pi*n, 4pi/3 +2pi*n, where n is an integer.
:)