How do you solve #2(sin(x/2))^2 + (cosx)^2 = 3#?

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Answer 1 · 2017-01-01T20:49:16

Please see the explanation.

Given: #2(sin(x/2))^2 + (cos(x))^2 = 3#

Here is a reference for Half Angle Formulas .

Substitute #sqrt((1 - cos(x))/2)# for #sin(x/2)#:

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

The square of the square root makes them both disappear:

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

The 2s cancel:

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

Reorder into quadratic form:

#(cos(x))^2 - cos(x) - 2 = 0#

It factors:

#(cos(x) - 2)(cos(x) + 1) = 0#

#cos(x) = 2 and cos(x) = -1#

Discard #cos(x) = 2# because it is out of range.

#cos(x) = -1#

#x = pi + 2npi#

where n is any negative or positive integer, including 0.