How do youFind all the solutions of the equation in the interval [0,2pi).? -5sinx = -2cos^2x + 4

-5sinx = -2cos^2x + 4

1 Answer
Jun 9, 2018

x=(7pi)/6, (11pi)/6

Explanation:

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-5sinx=-2cos^2x+4

-5sinx=-2(1-sin^2x)+4

-5sinx=-2+2sin^2x+4

2sin^2x+5sinx+2=0

sinx=(-5+-sqrt(25-4(2)(2)))/4=(-5+-3)/4=-2,-1/2

sinx=-2 is invalid because sin values are always between 1 and -1.

sinx=-1/2, :. x=(7pi)/6, (11pi)/6