How do you solve #4sin^2 x - 3cosx -3 = 0#?

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Answer 1 · 2015-04-23T17:41:55

We would rather have just one trigonometric function if possible.
How are #sinx# and #cosx# (the two functions we see here) related?

Yes, #sin^2x+cos^2x=1#, so we can replace #sin^2x# with #1-cos^2x#.

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

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

So we need to solve:

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

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

Solve for #cosx# first,
then for #x#