# How to solve equation cos 3x-10sin^2x=3cosx-10?

*print*Print*list*Cite

### 2 Answers

Solve `cos(3x)-10sin^2(x)=3cos(x)-10` :

Write 3x as 2x+x, and use the pythagorean identity

`cos(2x+x)-10(1-cos^2x)=3cosx-10`

** `cos(A+B)=cosAcosB-sinAsinB`

`cos2xcosx-sin2xsinx-10+10cos^2x=3cosx-10`

** Choose `cos2x=1-2sin^2x` ,and use `sin2x=2sinxcosx`

`(1-2sin^2x)(cosx)-2sin^2xcosx+10cos^2x=3cosx`

`cosx-2sin^2xcosx-2sin^2xcosx+10cos^2x=3cosx`

** Use pythagorean identity; subtract cosx from each side

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

`-4cosx+4cos^3x+10cos^2x=2cosx`

`4cos^3x+10cos^2x-6cosx=0`

`2cosx(2cos^2x+5cosx-3)=0`

`2cosx(2cosx-1)(cosx+3)=0`

By the zero product property (and assuming we are in the real numbers) we have either :

`2cosx=0,2cosx-1=0,cosx+3=0`

`cosx !=-3` for any real x

`2cosx=0==> cosx=0 ==>x=pi/2+npi` for `n in ZZ`

`2cosx-1=0 ==> cosx=1/2==>x=pi/3+2npi` or `x=2npi-pi/3` for `n in ZZ`

------------------------------------------------------------------

The solutions are `x=pi/2+npi,2npi +- pi/3`

-----------------------------------------------------------------

**Sources:**

You should write all the equation in terms of cos x such that:

`(4cos^3 x - 3cos x) - 10(1 - cos^2 x) = 3 cos x - 10`

Notice that `4cos^3 x - 3cos x ` substitutes `cos^3 x` and `1 - cos^2 x` substitutes `sin^2 x.`

Opening the brackets yields:

`4cos^3 x - 3cos x - 10 + 10cos^2 x = 3 cos x - 10`

You need to reduce like terms such that:

`4cos^3 x - 3cos x + 10cos^2 x = 3 cos x`

You should move all the terms to the left side such that:

`4cos^3 x + 10cos^2 x - 6 cos x= 0`

You should factor out `2 cos x` such that:

`2 cos x(2cos^2 x + 5 cos x - 3) = 0`

You should solve the equations `2 cos x = 0` and `2cos^2 x + 5 cos x - 3 = 0` such that:

`2 cos x = 0 =gt cos x = 0 =gt x = pi/2 + 2npi`

You need to solve the equation `2cos^2 x + 5 cos x - 3 = 0` substituting y for cos x such that:

`2y^2 + 5y - 3 = 0`

You need to use quadratic formula such that:

`y_(1,2) = (-5+-sqrt(25 + 24))/4`

`y_(1,2) = (-5+-sqrt49)/4`

`y_(1,2) = (-5+-7)/4 =gt y_1 = 1/2 ; y_2 = -3`

You need to solve for x the equations `cos x = 1/2` and `cos x = -3` .

Notice that the second equation is impossible to be solved since the values of cosine are not smaller than -1.

Hence, you only should solve `cos x = 1/2` such that:

`cos x = 1/2 =gt x = +-cos^(-1)(1/2) + 2npi`

`x = +-(pi/3) + 2npi`

**Hence, evaluating the solutions to the given equation yields`x = +-(pi/3)` `+ 2npi ` and `x = (pi/2) + 2npi` .**

**Sources:**