How do you find all the solutions of the following equation in the interval [0, 2pi)? 48sin^2x = 48-24cosx

1 Answer
Teddy
Jun 22, 2018

x={pi/2,(3pi)/2,pi/3,(5pi)/3}x={π2,3π2,π3,5π3}

Explanation:

48sin^2x=48-24cosx48sin2x=48−24cosx
2sin^2x=2-cosx2sin2x=2−cosx
2(1-cos^2x)=2-cosx2(1−cos2x)=2−cosx
2-2cos^2x=2-cosx2−2cos2x=2−cosx
0=2cos^2x-cosx0=2cos2x−cosx
0=cosx(2cosx-1)0=cosx(2cosx−1)
cosx=0cosx=0 or 2cosx-1=02cosx−1=0
x=pi/2,(3pi)/2x=π2,3π2 or x=pi/3,(5pi)/3x=π3,5π3

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