If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$, then the value of $x^{2025}+x^{2026}+x^{2027}$ is
If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$, then the value of $x^{2025}+x^{2026}+x^{2027}$ is
- $-1$
- $0$
- $1$
- $3$
Solution
$\begin{aligned} & \cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi \\ & \text { Since } 0 \leq \cos ^{-1} x \leq \pi \\ & \\ & 0 \leq \cos ^{-1} y \leq \pi \text { and } 0 \leq \cos ^{-1} z \leq \pi \\ & \\ & \text { Here, } \cos ^{-1} x=\cos ^{-1} y=\cos ^{-1} z=\pi \\ & \Rightarrow x=y=z=\cos \pi=-1 \\ & \therefore \quad x^{2005}+x^{2026}+x^{2027} \\ & =(-1)^{2025}+(-1)^{2026}+(-1)^{2027} \\ & =-1+1-1 \\ & =-1\end{aligned}$
Asked in: MHT CET 2023 (14 May Shift 1)
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