Let S = x: ^-1x= + ^-1x+ ^-1[2x+1].

Question

Let S = ${x:{\cos }^{-1}x=\pi +{\sin }^{-1}x+{\sin }^{-1}[2x+1]}$. Then $\sum _{x\in S}(2x-1{)}^{2}$ is equal to \_\_\_\_\_\_\_.

Answer: 5

Accepted Answer

Let $\alpha=\cos^{-1}x$, $\beta=\sin^{-1}(2x+1)$. Since $\sin^{-1}x=\dfrac{\pi}{2}-\alpha$, the equation becomes

$$\alpha=\pi+\left(\frac{\pi}{2}-\alpha\right)+\beta\ \Rightarrow\ 2\alpha-\beta=\frac{3\pi}{2}\ \Rightarrow\ 2\alpha=\frac{3\pi}{2}+\beta.$$

Taking cosine: $\cos2\alpha=\cos\left(\dfrac{3\pi}{2}+\beta\right)=\sin\beta$, i.e.

$$2\cos^2\alpha-1=\sin\beta\ \Rightarrow\ 2x^2-1=2x+1\ \Rightarrow\ x^2-x-1=0\ \Rightarrow\ x=\frac{1\pm\sqrt5}{2}.$$

Checking the domain of $\sin^{-1}(2x+1)$ (needs $2x+1\in[-1,1]$, i.e. $x\in[-1,0]$), only $x=\dfrac{1-\sqrt5}{2}$ is admissible; $x=\dfrac{1+\sqrt5}{2}$ is rejected.

For this root, $x^2-x=1\Rightarrow 4x^2-4x=4\Rightarrow(2x-1)^2=4x^2-4x+1=5$.

$$\sum_{x\in S}(2x-1)^2=5.$$

Solutions · 1