# Solve `3costhetasin^2theta - costheta = 0` for the interval `0<=theta<=Pi` ` ` ` `

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### 2 Answers

The first step is modifying the equation to be composed of one trinometric function.

We know that `sin^2theta=1-cos^2theta` :

`3costheta(1-cos^2theta)-costheta=0`

Expand and simplify:

`3costheta-3cos^3theta-costheta=0`

`costheta(3-3cos^2theta-1)=0`

`costheta(2-3cos^2theta)=0`

On the interval `0lt=thetalt=pi` :

`costheta=0 -gt theta=pi/2`

`2-3cos^2theta=0`

`costheta=sqrt(2/3)-gttheta=0.615`

Therefore, the roots of this equation are `pi/2 ` rad and 0.615 rad.

**Sources:**

We have given

`3cos(theta)sin^2(theta)-cos(theta)=0`

`and 0<=theta<=pi`

`cos(theta)(3sin^2(theta)-1)=0`

`either `

`cos(theta)=0`

`theta=pi/2 in (0,pi)`

`or`

`3sin^2(theta)-1=0`

`sin^2(theta)=1/3`

`sin(theta)=sqrt(1/3)`

`sin(theta)=-sqrt(1/3)` not possible because `theta in(0,pi)`

Thus

`sin(theta)=sqrt(1/3)=sin(35.26^o)`

`theta=35.26^o=pi/5 rad` (approx)

Thus our answers are `theta=pi/5 ,pi/2 rad.`