If xϵ[0,π] and 3sin2x+2cos2x+31−sin2x+2sin2x=28, then find number of values of x satisfy given equation.
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Solution
3sin2x+2sin2x+31−sin2x+2sin2x=28⇒3sin2x+2cos2x+33−(sin2x+2cos2x)=28 Let y=3sin2x+2cos2x ∴y=27y=28 ⇒y2−28y+27=0⇒y=27 or 1 If y=27 then 3sin2x+2cos2x=33 ⇒sin2x+(2cos2x−1)=0 ⇒sin2x+cos2x=2 Dividing both sides by √2 sin(2x+π4)=√2 which is not possible Now if y=1 ⇒3sin2x+2cos2x=1=30⇒sin2x+2cos2x=0⇒2cosx(sinx+cosx)=0 ∴cosx=0 or tanx=−1