How do you solve #4sin^2x=2cosx+1# and find all solutions in the interval #[0,2pi)#?

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Answer 1 · 2018-03-19T19:10:13

There are two solutions, #cos^-1((sqrt(13)-1)/4)#, and #2pi-cos^-1((sqrt(13)-1)/4)#

Use the Pythagorean Theorem to replace #sin^2x# with #1-cos^2x#.

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

Use the distributive property

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

Put this "quadratic" equation in standard form.

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

Use the quadratic formula to solve for #cosx#.

#cosx=(-2+-sqrt(2^2-4*4*3))/(2*4)=(-2+-sqrt(52))/(2*4)=(-1+-sqrt(13))/(4)#

We can throw out #(-1-sqrt(13))/4# as a real root because it is less than -1, and #cosx# cannot be less than -1 if #x# is real.

Therefore we need only find angles between 0 and #2pi# whose cosine are #(sqrt(13)-1)/4#. These angle would be

#x=cos^-1((sqrt(13)-1)/4)# and #x=2pi-cos^-1((sqrt(13)-1)/4)#.