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.

How do you solve 2(sin(x/2))^2 + (cosx)^2 = 32(sin(x/2))^2 + (cosx)^2 = 3?