How do you solve #cos 2x- sin^2 (x/2) + 3/4 = 0#?

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Answer 1 · 2015-08-10T15:06:12

#Cosx=1/2# and #cosx=-3/4#

Step 1:
#cos2x-Sin^2(x/2)+3/4=0#
Use #cos2x=cos^2x-sin^2x#

Step 2:
#cos^2x-sin^2x-sin^2(x/2)+3/4=0#
Use #sin^2x+cos^2x=1#

Step3:
#2cos^2x-1-sin^2(x/2)+3/4=0#
Use #cosx=1-2sin^2(x/2)# (Double angle formula).

Step 4:
#2cos^2x-1-1/2+1/2cosx+3/4=0#
#2cos^2x+2cosx-3=0#

Multiply by 4 to get

#8cos^x+2cosx-3=0#

Step 5: Solve the quadratic equation to get

#(2cos-1)(4cosx+3)=0#
#cosx=1/2# and #cosx=-3/4#