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Solve the equation `2sin^2x-5cos(pi/2-x)+2=0` for x

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lemong | Student, Undergraduate | (Level 1) Honors

Posted July 1, 2012 at 2:13 PM via web

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Solve the equation `2sin^2x-5cos(pi/2-x)+2=0` for x

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted July 1, 2012 at 2:20 PM (Answer #1)

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The equation `2sin^2x-5cos(pi/2-x)+2=0` has to be solved.

`2sin^2x-5cos(pi/2-x)+2=0`

=> `2sin^2x-5*sin x+2=0`

=> `2sin^2x-4*sin x - sin x + 2 = 0`

=> `2*sin x(sin x - 2) - 1(sin x - 2) = 0`

=> `(2*sin x - 1)(sin x - 2) = 0`

=> `sin x = 1/2` and `sin x = 2`

As sin x lies in [-1, 1] the root sin x = 2 can be ignored.

This gives sin x = 1/2.

`x = sin^-1(1/2)`

=> `x = pi/6 + 2*n*pi` and `x = 5*pi/6 + 2*n*pi`

The solution of the equation is `x = pi/6 + 2*n*pi` and `x = 5*pi/6 + 2*n*pi`

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