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Question #7ee2f

1 Answer

Aug 13, 2017

The solutions are #S={0, pi, 7/6pi, 11/6pi}#

Explanation:

We need

#cos^2x+sin^2x=1#

Our equation is

#2cos^2x=2+sinx#

Therefore,

#2(1-sin^2x)=2+sinx#

#2-2sin^2x=2+sinx#

#2sin^2x+sinx=0#

Factorising

#sinx(2sinx+1)=0#

so,

#sinx=0#, #=>#, #x=0# or #x=pi#

#sinx=-1/2#, #=>#, #x=7/6pi# or #x=11/6pi#

#AA x in [0, 2pi)#

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