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}#

Explanation:

#48sin^2x=48-24cosx#
#2sin^2x=2-cosx#
#2(1-cos^2x)=2-cosx#
#2-2cos^2x=2-cosx#
#0=2cos^2x-cosx#
#0=cosx(2cosx-1)#
#cosx=0# or #2cosx-1=0#
#x=pi/2,(3pi)/2# or #x=pi/3,(5pi)/3#

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