Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy sin^-1( 3x 5)+ sin^-1(…

$${\mathrm{sin}}^{-1}\frac{3x}{5}+{\mathrm{sin}}^{-1}\frac{4x}{5}={\mathrm{sin}}^{-1}x$$sin−153x​+sin−154x​=sin−1x

$${\mathrm{sin}}^{-1}(\frac{3x}{5}\sqrt{1-\frac{16{x}^{2}}{25}}+\frac{4x}{5}\sqrt{1-\frac{9{x}^{2}}{25}})={\mathrm{sin}}^{-1}x$$sin−1(53x​1−2516x2​​+54x​1−259x2​​)=sin−1x

$$\frac{3x}{5}\sqrt{1-\frac{16{x}^{2}}{25}}+\frac{4x}{5}\sqrt{1-\frac{9{x}^{2}}{25}}=x$$53x​1−2516x2​​+54x​1−259x2​​=x

$$x=0,3\sqrt{25-16{x}^{2}}+4\sqrt{25-9{x}^{2}}=25$$x=0,325−16x2​+425−9x2​=25

$$4\sqrt{25-9{x}^{2}}=25-3\sqrt{25-16{x}^{2}}$$425−9x2​=25−325−16x2​

squaring we get

$$16(25-9{x}^{2})=625+9(25-16{x}^{2})-150\sqrt{25-16{x}^{2}}$$16(25−9x2)=625+9(25−16x2)−15025−16x2​

$$400=625+225-150\sqrt{25-16{x}^{2}}$$400=625+225−15025−16x2​

$$\sqrt{25-16{x}^{2}}=3\Rightarrow 25-16{x}^{2}=9$$25−16x2​=3⇒25−16x2=9

$$\Rightarrow {x}^{2}=1$$⇒x2=1

Put $x=0,1,-1$x=0,1,−1 in the original equation We see that all values satisfy the original equation.

Number of solution $=3$=3