Question #28bc1

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Answer 1 · 2017-03-26T09:16:36

$2 {sin}^{- 1} x - 3 {cos}^{- 1} x = \frac{π}{6}$ $2sin^-1x-3cos^-1x=pi/6$

Putting ${cos}^{- 1} x = \frac{π}{2} - {sin}^{- 1} x$ $color(blue)(cos^-1x=pi/2-sin^-1x)$

$\Rightarrow 2 {sin}^{- 1} x - 3 (\frac{π}{2} - {sin}^{- 1} x) = \frac{π}{6}$ $=>2sin^-1x-3(pi/2-sin^-1x)=pi/6$

$\Rightarrow 2 {sin}^{- 1} x - \frac{3 π}{2} + 3 {sin}^{- 1} x = \frac{π}{6}$ $=>2sin^-1x-(3pi)/2+3sin^-1x=pi/6$

$\Rightarrow 5 {sin}^{- 1} x = \frac{3 π}{2} + \frac{π}{6} = \frac{10 π}{6} = \frac{5 π}{3}$ $=>5sin^-1x=(3pi)/2+pi/6=(10pi)/6=(5pi)/3$

$\Rightarrow 5 {sin}^{- 1} x = \frac{5 π}{3}$ $=>5sin^-1x=(5pi)/3$

$\Rightarrow {sin}^{- 1} x = \frac{π}{3}$ $=>sin^-1x=pi/3$

$\Rightarrow x = sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ $=>x=sin(pi/3)=sqrt3/2$

please note

Let ${sin}^{- 1} x = θ$ $sin^-1x=theta$

$\Rightarrow sin θ = x$ $=>sintheta=x$

$\Rightarrow cos (\frac{π}{2} - θ) = x$ $=>cos(pi/2-theta)=x$

$\Rightarrow \frac{π}{2} - θ = {cos}^{- 1} x$ $=>pi/2-theta=cos^-1x$

$\Rightarrow \frac{π}{2} - {sin}^{- 1} x = {cos}^{- 1} x$ $=>pi/2-sin^-1x=cos^-1x$