If xϵ[0,π] and 81sin2x+81cos2x=30, then find number of values of x satisfy the given equation.
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Solution
Let 81sin2x=y 81cos2x=811−sin2x=81y−1 So that the given equation can be written as y2−30y+81=0⇒y=3 or y=27 ⇒81sin2x=3 or 27 ⇒34sin2x=31 or 33 ⇒4sin2x=1 or 3 ⇒sin2x=14 or 34 ⇒sinx=±12 or ±√32 ⇒x=π6,π3