How do you solve 2cos2x-3sinx=1?

2 Answers
Dean R.
May 1, 2018

x = arcsin(1/4) + 360^circ k or

x=(180^circ - arcsin(1/4)) + 360^circ k or

x = -90^circ + 360^circ k for integer k.

Explanation:

2 cos 2x - 3 sin x = 1

The useful double angle formula for cosine here is

cos 2x = 1 - 2 sin ^2 x

2(1 - 2 sin ^2 x) - 3 sin x = 1

0 = 4 sin ^2 x + 3 sin x - 1

0 = (4 sin x - 1 )( sin x + 1)

sin x = 1/4 or sin x = -1

x = arcsin(1/4) + 360^circ k or x=(180^circ - arcsin(1/4)) + 360^circ k or x = -90^circ + 360^circ k for integer k.

Abhishek K.
May 1, 2018

rarrx=npi+(-1)^n*sin^(-1)(1/4) or npi+(-1)^n*(-pi/2) nrarrZ

Explanation:

rarr2cos2x-3sinx-1=0

rarr2(1-2sin^2x)-3sinx-1=0

rarr2-4sin^2x-3sinx-1=0

rarr4sin^2x+3sinx-1=0

rarr(2sinx)^2+2*(2sinx)*(3/4)+(3/4)^2-(3/4)^2-1=0

rarr(2sinx+3/4)^2=1+9/16=25/16

rarr2sinx+3/4=+-sqrt(25/16)=+-(5)/4

rarr2sinx=+-5/4-3/4=(+-5-3)/4

rarrsinx=(+-5-3)/8

Taking +ve sign, we get

rarrsinx=(5-3)/8=1/4

rarrx=npi+(-1)^n*sin^(-1)(1/4) nrarrZ

Taking -ve sign, we get

rarrsinx=(-5-3)/8=-1

rarrx=npi+(-1)^n*(-pi/2) where nrarrZ

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