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

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claymore | Student, Undergraduate | (Level 2) eNoter

Posted July 5, 2012 at 4:25 AM via web

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

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted July 5, 2012 at 4:36 AM (Answer #1)

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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.

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