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

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Answer 1 · 2018-05-01T02:15:09

# 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#.

#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#.

Answer 2 · 2018-05-01T02:20:15

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

#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#