How do you solve #2cos^2x+3sinx=3# in the interval #0<=x<=2pi#?

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How do you solve #2cos^2x+3sinx=3# in the interval #0<=x<=2pi#?

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Answer 1 · 2016-09-26T11:16:48

Solution # x= pi/6,pi/2, (5pi)/6# for # 0 <=x<=2pi#

Explanation:

#2cos^2x+3sinx=3 or 2(1-sin^2x)+3sinx-3=0 or 2sin^2x-3sinx+1=0 or (2sinx-1)(sinx-1)=0#. So either #2sinx-1=0 or sinx-1=0 # When #2sinx-1=0 :.2sinx=1 :. sinx= 1/2; sin(pi/6)=1/2 and sin(pi-pi/6)=1/2 :. x=pi/6,(5pi)/6# When #sinx-1=0 or sinx=1 ; sin (pi/2)=1 :. x= pi/2#
Solution # x= pi/6,pi/2, (5pi)/6# for # 0 <=x<=2pi# [Ans]

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How do you solve #2cos^2x+3sinx=3# in the interval #0<=x<=2pi#?

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