The number of values of such that is
This problem is part of the set Trigonometry .
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Let u = s i n ( x ) so that 2 u 2 + 5 u − 3 = ( 2 u − 1 ) ( u + 3 ) = 0 ⇒ u = 2 1 , − 3 . Since − 1 ≤ s i n ( x ) ≤ 1 , only s i n ( x ) = 2 1 is permissible. For the interval x ∈ [ 0 , 3 π ] , there are only 4 possible solutions: x = 6 π , 6 5 π , 6 1 3 π , 6 1 7 π .