How do you solve #3\sin ^ { 2} x + 4\cos x - 4= 0#?

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Answer 1 · 2018-02-01T04:18:30

The answers are #x = cos^-1(1/3) + 2pik, 2pik#.

#3sin^2x+4cosx-4=0#

Use the identity #sin^2x+cos^2x=1 => sin^2x=1-cos^2x#

#=>3(1-cos^2x)+4cosx-4=0#

#3-3cos^2x+4cosx-4=0#

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

Let #u = cosx#:

#-1-3u^2+4u=0#

#-3u^2+4u-1=0#

#3u^2-4u+1=0#

#(3u-1)(u-1)=0#

#u=1/3, 1#

Now plug #cosx# back in for #u#:

#cosx=1/3, cosx=1#

#x = cos^-1(1/3) + 2pik, 2pik#

where #k# is any integer. For example, #x# could be #0#, #2pi#, #4pi#, etc.