Question

sin-1(4x)+sin-1(3x)=-π\2​

Answer 1

Inverse Circular Function

Formula:

$sin^{-1}x+sin^{-1}y=sin^{-1}[x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}]$

Now we solve the problem:

$\quad sin^{-1}(4x)+sin^{-1}(3x)=-\frac{\pi}{2}$

$\to sin^{-1}[4x\sqrt{1-9x^{2}}+3x\sqrt{1-16x^{2}}]=-\frac{\pi}{2}$

$\to 4x\sqrt{1-9x^{2}}+3x\sqrt{1-16x^{2}}=sin(-\frac{\pi}{2})=-1$

$\to 4x\sqrt{1-9x^{2}}+1=-3x\sqrt{1-16x^{2}}$

Squaring both sides, we get:

$\quad 16x^{2}(1-9x^{2})+8x\sqrt{1-9x^{2}}+1=9x^{2}(1-16x^{2})$

$\to 16x^{2}-144x^{4}+8x\sqrt{1-9x^{2}}+1=9x^{2}-144x^{4}$

$\to 8x\sqrt{1-9x^{2}}=-7x^{2}-1$

Squaring both sides, we get:

$\quad 64x^{2}(1-9x^{2})=49x^{4}-14x^{2}+1$

$\to 64x^{2}-576x^{4}=49x^{4}-14x^{2}+1$

$\to 625x^{4}-50x^{2}+1=0$

$\to (25x^{2}-1)^{2}=0$

Either $25x^{2}-1=0$ or, $25x^{2}-1=0$

$\implies \boxed{x=\pm \frac{1}{5},\:\pm \frac{1}{5}}$

This is the required solution.

Answer 2

Answer:

sin^{-1}x+sin^{-1}y=sin^{-1}[x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}]

Now we solve the problem:

\quad sin^{-1}(4x)+sin^{-1}(3x)=-\frac{\pi}{2}

\to sin^{-1}[4x\sqrt{1-9x^{2}}+3x\sqrt{1-16x^{2}}]=-\frac{\pi}{2}

\to 4x\sqrt{1-9x^{2}}+3x\sqrt{1-16x^{2}}=sin(-\frac{\pi}{2})=-1

\to 4x\sqrt{1-9x^{2}}+1=-3x\sqrt{1-16x^{2}}

Squaring both sides, we get:

\quad 16x^{2}(1-9x^{2})+8x\sqrt{1-9x^{2}}+1=9x^{2}(1-16x^{2})

\to 16x^{2}-144x^{4}+8x\sqrt{1-9x^{2}}+1=9x^{2}-144x^{4}

\to 8x\sqrt{1-9x^{2}}=-7x^{2}-1

Squaring both sides, we get:

\quad 64x^{2}(1-9x^{2})=49x^{4}-14x^{2}+1

\to 64x^{2}-576x^{4}=49x^{4}-14x^{2}+1

\to 625x^{4}-50x^{2}+1=0

\to (25x^{2}-1)^{2}=0

Either 25x^{2}-1=0 or, 25x^{2}-1=0

\implies \boxed{x=\pm \frac{1}{5},\:\pm \frac{1}{5}}

This is the required solution.

Step-by-step explanation:

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