# Solve trigonometric equation `cos^2 2x + cos^2 4x=1` ?

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### 2 Answers

we know `cos2x = 2cos^2x-1`

Then `cos^2x = (1+cos2x)/2`

Lets assume that x=2x

Then

`cos^2(2x) =(1+cos(2*2x))/2 = (1+cos4x)/2`

`cos^2 2x + cos^2 4x=1`

`(1+cos4x)/2+cos^2 (4x) = 1`

`1+cos4x+2cos^2(4x) =2`

`2cos^2(4x)+cos4x-1 = 0`

`2cos^2(4x)+2cos4x-cos4x-1 = 0`

`2cos4x(cos4x+1)-1(cos4x+1) = 0`

`(2cos4x-1)(cos4x+1) =0`

`2cos4x-1 = 0 `

`cos4x = 1/2`

`cos4x=cos(pi/3)`

`4x = 2n(pi)+-(pi/3)` using general solution for cosins

`x = n(pi/2)+-(pi/12)`

`cos4x+1 = 0`

`cos4x = -1`

`cos4x = cospi`

`4x = 2m(pi)+-pi` using general solution for cosins

`x = m(pi/2)+-pi/4`

**So the solutions for x is ;**

`x= n(pi/2)+-(pi/12)`

**or**

**`x = m(pi/2)+pi/4` **

**` `****` ` where `n,m in Z` **

**Sources:**

The equation `cos^2 2x + cos^2 4x = 1` has to be solved.

`cos 4x = 2*cos^2 2x - 1`

=> `cos^2x = (cos 4x + 1)/2`

`cos^2 2x + cos^2 4x = 1`

=> `(cos 4x + 1)/2 + cos^2 4x = 1`

Let `cos 4x = y`

=> `y/2 + 1/2 + y^2 = 1`

=> `y + 1 + 2y^2 = 2`

=> `2y^2 + y - 1 = 0`

=> `2y^2 + 2y - y - 1 = 0`

=> `2y(y + 1) - 1(y + 1) = 0`

=> `(2y - 1)(y + 1) = 0`

=> `y = 1/2` and `y = -1`

`cos 4x = 1/2` and `cos 4x = -1`

=> `4x = 60` , `4x = 300` , `4x = 180`

=> x = 15, x = 75, x = 45

**The solution of the equation is x = 15 + 360*n, x = 45 + n*360 and x = 75 + n*360 degrees.**