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Cos x = sin 2x value of x |
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Answer» ong>ANSWER: We need sin2x=2sinxcosx sin2x =cosxsin2x−COSX=0 2sinxcosx−cosx=0 cosx(2sinx−1)=0 So, {cosx=0 2sinx−1=0⇔ , { cos x = 0 sin x = 1 2
⇔ , { x = π 2 3 2 π x = 1 6 π 5 6 π
∀ x ∈ [ 0 , 2 π ]
The solutions are S = { 1 2 π , 3 2 π , 1 6 π , 5 6 π }
GRAPH{sin(2x)-cosx [-1.622, 9.475, -2.51, 3.04]} |
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