Solve 2cos^2x+sinx=1 ?

1 Answer
Cesareo R.
Apr 14, 2017

x = (pi/2+2k pi uu -pi/6 +2kpi) for k in NN

Explanation:

Using the identity

sin^2x+cos^2x=1 and substituting

2(1-sin^2x)+sinx-1=0 or

2sin^2x-sinx-1=0

now solving for sinx

sinx = (1pm sqrt(1+8))/4 = {(1),(-1/2):}

then

x = (pi/2+2k pi uu -pi/6 +2kpi) for k in NN

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