Question

Solve the following equation for x: $sin^{-1}(1 - x) - 2sin^{-1}x = \frac{\pi}{2}$.

Answer 1

As we know that

$2 {sin}^{ - 1} x = {sin}^{ - 1} (2x \sqrt{1 - {x}^{2} } )$
and
${sin}^{ - 1} x + {sin}^{ - 1}y = {sin}^{ - 1} (x \sqrt{1 - {y}^{2} } + y \sqrt{1 - {x}^{2} } ) \\ \\ {sin}^{ - 1} (1 - x) + {sin}^{ - 1}(2x \sqrt{1 - {x}^{2} }) = {sin}^{ - 1} ((1 - x) \sqrt{1 - 4 {x}^{2} (1 - {x}^{2})} + (2x \sqrt{1 - {x}^{2} }) \sqrt{1 - {x}^{2} } )$
so
$((1 - x) \sqrt{1 - 4 {x}^{2} (1 - {x}^{2})} + (2x \sqrt{1 - {x}^{2} }) \sqrt{1 - {x}^{2} } ) = sin \: ( \frac{\pi}{2} ) \\ \\ ((1 - x) \sqrt{1 - 4 {x}^{2} (1 - {x}^{2})} + (2x \sqrt{1 - {x}^{2} }) \sqrt{1 - {x}^{2} } ) =1 \\ \\ 2x(1 - {x}^{2} ) = 1 - (1 - x) \sqrt{1 - 4 {x}^{2} + 4 {x}^{2} } ) \\ \\ x(1 - {x}^{2} ) = 1 - (1 - x)(1 - 2x) \\ \\ x - {x}^{3} = 1 - 1 + 2x + x - 2 {x}^{2} \\ \\ x - {x}^{3} = 3x - 2 {x}^{2} \\ \\ - {x}^{3} + 2 {x}^{2} - 2x =0 \\ \\ {x}^{3} - 2 {x}^{2} + 2x =0$
$x{(x}^{2} - 2 {x} + 2) =0 \\ \\ x = 0 \\ \\ (x^{2} - 2 {x} + 2) = 0 \\ \\ x_{1,2} = \frac{2 ± \sqrt{4 - 8} }{2} \\ \\ x_{1,2} = 1 ±i \\ \\ x_{1} = 1 + i \\ \\ x_{2} = 1 - i$

Answer 2

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18 ANSWERS



Pramod Madhukar

Answered May 19, 2018

cos^-1x+sin^-1(x/2)=pie/6

we know cos^-1(x/2)+sin^-1(x/2)=pie/2

so sin^-1(x/2)=pie/2-cos^-1(x/2)

substitute we get

cos^-1(x)-cos^-1(x/2)=pie/6-pie/2

cos^-1(x)-cos^-1(x/2)=-pie/3

cos^-1(x)-cos^-1(x/2)=-cos^-1(1/2)

cos^-1(x)=cos^-1(x/2)-cos^-1(1/2)

A=cos^-1(x/2)

B=cos^-1(1/2)

cosA=x/2

sinA=sqrt(4-x^2)/2

cosB=1/2

sinB=sqrt(3)/2

cos^-1(x)=A-B

x=cos(A-B)

cosAcosB+sinAsinB=x

(x/2)(1/2)+sqrt(4-x^2)/2 * sqrt(3)/2

x/4 +sqrt(3)/4 sqrt(4-x^2)

x-x/4=(3)^1/2/4 *sqrt(4-x^2)

x(3/4)=3^1/2 /4 * sqrt(4-x^2)

3x=root 3 *sqrt(4-x^2)

9x^2=12–3x^2

x^2=1

x=+1 ,x=-1