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Class-11-commerce»Maths

solve
cos3x = sin2x

To solve \cos 3 x = \sin 2 x Consider the domain 0 < x < π take \cos 3 x = \cos (2 x + x) = \cos 2 x cosx - \sin 2 x sinx (since \cos (A + B) = cosAcosB - sinAsinB) = (2 \cos^{2} x - 1) cosx - 2 sinx . cosx . sinx = 2 \cos^{3} x - cosx - 2 cosx (1 - \cos^{2} x) = 2 \cos^{3} x - cosx - 2 cosx + 2 \cos^{3} x = 4 \cos^{3} x - 3 cosx = cosx (4 \cos^{2} x - 3) = cosx (4 (1 - \sin^{2} x) - 3) = cosx (1 - 4 \sin^{2} x) so, cosx (1 - 4 \sin^{2} x) = \sin 2 x cosx (1 - 4 \sin^{2} x) = 2 sinxcosx cosx (4 \sin^{2} x + 2 sinx - 1) = 0 therefore, cosx = 0 so, x = \frac{π}{2} and 4 \sin^{2} x + 2 sinx - 1 = 0 so, sinx = \frac{- 2 \pm \sqrt{4 - 4 (4) (- 1)}}{8} = \frac{- 2 \pm \sqrt{20}}{8} sinx = \frac{- 2 \pm 2 \sqrt{5}}{8} = \frac{- 1 \pm \sqrt{5}}{4} so, x = \sin^{- 1} (\frac{- 1 \pm \sqrt{5}}{4}) therefore values of x are \frac{π}{2}, \sin^{- 1} (\frac{- 1 \pm \sqrt{5}}{4}) .

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