Solve: `sqrt 3*cot x*sin x + 2 cos^2 x = 0`   for `0<= x<= 180`

1 Answer | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The equation `sqrt 3*cot x*sin x + 2cos^2 x = 0` has to be solved for `0 <= x<= 180` .

`sqrt 3*cot x*sin x + 2cos^2 x = 0`

=> `sqrt 3*(cos x/sin x)*sin x + 2cos^2 x = 0`

=> `sqrt 3*cos x + 2cos^2 x = 0`

=> `cos x(sqrt 3 + 2*cos x) = 0`

=> `cos x = 0` and `cos x = -sqrt 3/2`

=> x = 90 degrees and x = 150 degrees

The solutions of the equation in the given interval are x = 90 and x =150 degrees.

We’ve answered 315,526 questions. We can answer yours, too.

Ask a question