SOLUTION: Hi,
Can someone tell me the steps how to solve this
cox x + 1 = √3 sin x where 0 ≤ x < 2∏
the answer is cos x = 1/2 or cos x =-1
I just want to know how
Algebra.Com
Question 731700: Hi,
Can someone tell me the steps how to solve this
cox x + 1 = √3 sin x where 0 ≤ x < 2∏
the answer is cos x = 1/2 or cos x =-1
I just want to know how did they get it.
Thank you so much in advance.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
cos x + 1 = √3 sin x where 0 ≤ x < 2π
cos x + 1 = √3 sin x
square both sides
cos^2x+2cosx+1=3sin^2x
cos^2x+2cosx+1=3(1-cos^2x)
cos^2x+2cosx+1=3-3cos^2x)
4cos^2x+2cosx-2=0
(4cosx-2)(cosx+1)=0
..
4cosx-2=0
cosx=1/2
x=π/3, 5π/3 (In quadrants I and IV where cos>0)
or
cosx+1=0
cosx=-1
x=π
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