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

Asked on by mashymash

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crmhaske's profile pic

crmhaske | College Teacher | (Level 3) Associate Educator

Posted on

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

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


Expand and simplify:




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

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



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

pramodpandey's profile pic

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

We have given


`and 0<=theta<=pi`


`either `


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





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



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

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

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