# `1+(cos^2x)/2=1/2+sin^2 x` ``

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to start by bringing all the terms to a common denominator, such that:

`2 + cos^2 x = 1 + 2sin^2 x => 2 - 1 + cos^2 x = 2sin^2 x`

`1 + cos^2 x = 2 sin^2 x => 2sin^2 x - 1 - cos^2 x = 0`

Replacing `1 - sin^2 x` for `cos^2 x` yields:

`2sin^2 x - 1 - (1 - sin^2 x) = 0 => 3sin^2 x - 2 = 0 => sin^2 x = 2/3 => sin x = +-sqrt(2/3) => sin x = +-(sqrt6)/3 => x = (-1)^n*sin^(-1)(+-(sqrt6)/3) + n*pi`

Hence, evaluating x, under the given conditions, yields `x = (-1)^n*sin^(-1)(+-(sqrt6)/3) + n*pi.`

oldnick | (Level 1) Valedictorian

Posted on

Multiply both side by 2:

`2` `+` `cos^2 x` `=` `1` `+` `2sin^2x`

subract 1 both sides:

`2 -1+cos^2 x` `= 1 -1 + 2sin^2 x`

`1 + cos^2 x=` `2sin^2 x`

Knowing that:  `cos^2 x = 1 - sin^2 x`

`1 + 1 -sin^2 x` `=2sin^2 x`

`2 - sin^2 x = 2 sin^2 x`

`2 -sin^2 x + sin^2 x =2 sin^2 x +sin^2 x`

`2= 3 sin^2 x`

`2/3` `= sin^2 x`

using square root:

`sin x =` `+-`  `sqrt(2)/sqrt(3)` `=0,8164966`

`x = 54*44'08 ''`   `x=125* 15' 51''`

`x = 234* 44' 08''` `x= 305* 15' 51''`