solve 2sin²x + 3cos x = 0
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Changing sin
2
x into 1−cos
2
x; we get
2(1−cos
2
x)+
3
cosx+1=0
or 2cos
2
x−
3
cosx−3=0
∴cosx=
4
3
±
3+24
=
4
3
+3
3
=
3
or
2
−
3
Since
3
is greater than 1 it is not admissible as cosx can not be greater than 1.
∴cosx=−
3
/2=−cos(π/6)
=cos(π−π/6)=cos(5π/6)
∴x=2nπ±(5π/6).
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