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July 29, 2014

July 29, 2014

Posted by **Anonymous** on Wednesday, May 23, 2012 at 11:12pm.

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 -
**MathMate**, Thursday, May 24, 2012 at 8:07amUsing 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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