Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation cos^-1(x)-2 sin^-1(x)=…

Given,

$${\mathrm{cos}}^{-1}x-2{\mathrm{sin}}^{-1}x={\mathrm{cos}}^{-1}2x$$cos−1x−2sin−1x=cos−12x

$$\Rightarrow {\mathrm{cos}}^{-1}x-2(\frac{\pi }{2}-{\mathrm{cos}}^{-1}x)={\mathrm{cos}}^{-1}2x$$⇒cos−1x−2(2π​−cos−1x)=cos−12x

$$\Rightarrow {\mathrm{cos}}^{-1}x-\pi +2{\mathrm{cos}}^{-1}x={\mathrm{cos}}^{-1}2x$$⇒cos−1x−π+2cos−1x=cos−12x

$$\Rightarrow 3{\mathrm{cos}}^{-1}x=\pi +{\mathrm{cos}}^{-1}2x...(1)$$⇒3cos−1x=π+cos−12x...(1)

Taking $\mathrm{cos}$cos both side we get,

$$\Rightarrow \mathrm{cos}(3{\mathrm{cos}}^{-1}x)=\mathrm{cos}(\pi +{\mathrm{cos}}^{-1}2x)$$⇒cos(3cos−1x)=cos(π+cos−12x)

$$\Rightarrow \mathrm{cos}({\mathrm{cos}}^{-1}(4{x}^{3}-3x))=-\mathrm{cos}({\mathrm{cos}}^{-1}2x)$$⇒cos(cos−1(4x3−3x))=−cos(cos−12x)

$$\Rightarrow 4{x}^{3}-3x=-2x$$⇒4x3−3x=−2x

$$\Rightarrow 4{x}^{3}=x$$⇒4x3=x

$$\Rightarrow x=0,\pm \frac{1}{2}$$⇒x=0,±21​

All will satisfy the original equation,

So, sum of all the solution will be $0+\frac{1}{2}-\frac{1}{2}=0$0+21​−21​=0