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Mathematics Question on Trigonometric Equations

Solve the following equation sin−13x5+sin−14x5=sin−1 xsin^{-1} \frac{3x}{5} + sin^{-1} \frac{4x}{5} = sin^{-1} \,x

A

0,1,−10,1, -1

B

0,−10, -1

C

0,10,1

D

1,−11,-1

Answer

0,1,−10,1, -1

Explanation

Solution

We have, sin−13x5+sin−14x5=sin−1 xsin^{-1} \frac{3x}{5} + sin^{-1} \frac{4x}{5} = sin^{-1} \,x \Rightarrow sin^{-1}\left\\{\frac{3x}{5}\sqrt{1-\frac{16x^{2}}{25}}+\frac{4x}{5}\sqrt{1-\frac{9x^{2}}{25}}\right\\} = sin^{-1}\,x ⇒3x25−16x2+4x25−9x2=25x\Rightarrow 3x\sqrt{25-16x^{2}}+ 4x\sqrt{25-9x^{2}} = 25x ⇒x=0\Rightarrow x = 0 or, 325−16x2+425−9x2=253\sqrt{25-16x^{2}}+ 4\sqrt{25-9x^{2}} = 25 ⇒425−9x2=25−325−16x2\Rightarrow 4\sqrt{25-9x^{2}} = 25 - 3\sqrt{25 - 16x^{2}} Squaring both sides, we get 16(25−9x2)=625+9(25−16x2)−15025−16x216\left(25 - 9x^{2}\right) = 625 + 9\left(25 - 16x^{2}\right)-150\sqrt{25-16x^{2}} ⇒15025−16x2=450\Rightarrow 150\sqrt{25-16x^{2}} = 450 ⇒25−16x2=3\Rightarrow \sqrt{25-16x^{2} } = 3 Again squaring both sides, we get 25−16x2=925 - 16x^{2} = 9 ⇒x=±1\Rightarrow x = \pm 1 Hence, x=0x = 0, 11, −1-1 are the roots of the given equation.