# trig

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Solve cos x-1 = sin^2 x
Find all solutions on the interval [0,2pi)
a. x=pi, x=pi/2, x= 2pi/3
b. x=3pi/7, x=pi/2, x=2pi/3
c. x=3pi/7, x=3pi/2, x=3pi/2
d. x=pi, x=pi/2, x=3pi/2

• trig -

Using sin^2 x = 1-cos^2(x)
the equation becomes:
cos²(x)+cos(x)-2=0
Use the substitution c=cos(x) to transform the equation to :
c²+c-2=0
(c+2)(c-1)=0
c=-2 or c=+1
Since cos(x) cannot equal -2, solution is rejected.
Now solve for all values of 0≤x≤2π where cos(x)=1

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