# What is the general solution for the following and the values of `theta` within the domain `(-180; 180)` degrees: `cos(4theta-10)= -2cos2theta`

aruv | Student

Let us write question

`cos(4theta-10)=cos(4theta)cos(10)+sin(4theta)sin(10)`

`=(2cos^2(2theta)-1)cos(10)+2sin(2theta)cos(2theta) sin(10)`

`Let`

`cos(2theta)=x`

`then`

`cos(4theta-10)=(2x^2-1)cos(10)+2sqrt(1-x^2)xsin(10)`

`Thus`

`(2x^2-1)cos(10)+2xsqrt(1-x^2)sin(10)=-2x`

`(2x^2-1)cos(10)+2x=-2xsqrt(1-x^2)sin(10)`

`Square Both Side`

`(4x^4+1-4x^2)cos^2(10)+4x^2+4x(2x^2-1)cos(10)`

`=4x^2(1-x^2)sin^2(10)`

`4x^4 cos^2(10)+4x^4sin^2(10)+(-4x^2)(cos^2(10)+sin^2(10))`

`+cos^2(10)+4x^2+8x^3cos(10)-4x cos(10)=0`

`4x^4+8x^3cos(10)-4x^2+4x^2+cos^2(10)-4xcos(10)=0`

`4x^4+8x^3 cos(10)-4xcos(10)+cos^2(10)=0`

`4x^4+7.88x^3-3.94x+.97=0```

`x=.33,.40,-1.34+-.12i`

`cos(2theta)=.33=cos(70.73)`

`2theta=2nxx180+-70.73`

`theta=nxx180+-35.37`

`if n=0`

`theta=+-35.37`

`if n=1`

`theta=180-35.37=144.63`

`Now`

`cos(2theta)=.40=cos(66.42)`

`2theta=2mxx180+-66.42`

`theta=mxx180+-33.21`

`if m=0`

`theta=+-33.21`

`if m=1`

`theta=180-33.21=146.79`

`Thus`

`Ans`

`theta=+-35.37,+-33.21,144.63,146.79`

QED.