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Find the number of the solution of x if \cos [\cos^{-1}(\sin^2 x(2\cos^{-1}x))]=0

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

3


Answers (1)

best_answer

 

 

Multiple angles in terms of  arccos -

Multiple angles in terms of  arccos

 

\\\mathrm{1.\;\;\;\2\;cos^{-1}x=\left\{\begin{matrix} \cos ^{-1}\left(2 x^{2}-1\right), & &\text { if } 0 \leq x \leq 1\\ \\ 2\pi-\cos ^{-1}\left(2 x^{2}-1\right),& &\text { if }-1 \leq x \leq 0 \end{matrix}\right.}\\\\\\\mathrm{2.\;\;\;3\;\cos^{-1}x=\left\{\begin{matrix} \cos ^{-1}\left(4 x^{3}-3 x\right), & & \text { if } \frac{1}{2} \leq x \leq 1\\\\ 2\pi-\cos ^{-1}\left(4 x^{3}-3 x\right),& &\text { if }-\frac{1}{2} \leq x \leq \frac{1}{2} \\\\ 2\pi+\cos ^{-1}\left(4 x^{3}-3 x\right),& & \text { if }-1 \leq x \leq-\frac{1}{2} \end{matrix}\right.}

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\cos [\cos^{-1}(\sin^2 (2\cos^{-1}x))]=\sin [\cos^{-1}(\sin^2 (\cos^{-1} \{2x^2-1 \}))]\ \text{For }x>0 \\ \Rightarrow \cos [\cos^{-1}(1-\cos^2 (\cos^{-1} \{2x^2-1 \}))]=0\\ \Rightarrow \cos [\cos^{-1}(1-(\cos (\cos^{-1} \{2x^2-1 \})^2))]=0\\ \Rightarrow \cos [\cos^{-1}(1-(2x^2-1)^2)]=0\\ \Rightarrow \cos [\cos^{-1}(4x^2-4x^4)]=0\\ 4x^2-4x^4=0\\ x=\{0,1,\}\\ If\ x<0\\ \cos [\cos^{-1}(\sin^2 (2\cos^{-1}x))]=\cos [\cos^{-1}(\sin^2 (2\pi-\cos^{-1} \{2x^2-1 \}))]\ \\ \Rightarrow \cos [\cos^{-1}(1-\cos^2 (\cos^{-1} \{2x^2-1 \}))]=0\\ \Rightarrow \cos [\cos^{-1}(1-(\cos (\cos^{-1} \{2x^2-1 \})^2))]=0\\ \Rightarrow \cos [\cos^{-1}(1-(2x^2-1)^2)]=0\\ \Rightarrow \cos [\cos^{-1}(4x^2-4x^4)]=0\\ 4x^2-4x^4=0\\ x=-1\\ x=\{1,0,-1 \} \text{total 3 solutions }

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Sanket Gandhi

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