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

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

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Answer 1 · 2018-06-09T01:18:02

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

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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#