2cos^2x+sqrt(3)cosx=0 solution set: {pi/2, 3pi/2, 7pi/6, 5pi/6} I can't figure out how to get those solutions?

2 Answers
Oct 23, 2015

See the explanation below

Explanation:

The equation can be written as
cosx(2cosx+3)=0
which implies, either cosx=0or2cosx+3=0
If cosx=0 then the solutions are x=π2or3π2or(π2+nπ), where n is an integer
If 2cosx+3=0,thencosx=32,x=2π3+2nπor4π3+2nπ where n is an integer

Oct 23, 2015

Solve 2cos2x+3.cosx=0

Explanation:

cos x(2cos x + sqrt3) = 0
a. cos x = 0 --> x=π2 and x=3π2 (Trig unit circle)
b. cosx=32 --> x=±5π6 (Trig unit circle)
Note. The arc 5π6 is the same as the arc 7π6 (co-terminal)

Answers: π2;3π2;5π6and7π6