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Question: The equation \(2{{\cos }^{-1}}x+{{\sin }^{-1}}x=\dfrac{11\pi }{6}\) . (a) No solution. (b) Only ...

The equation 2cos⁡−1x+sin⁡−1x=11π62{{\cos }^{-1}}x+{{\sin }^{-1}}x=\dfrac{11\pi }{6} .
(a) No solution.
(b) Only one solution
(c) Two solutions
(d) Three solutions

Explanation

Solution

Hint: Start by using the identity cos⁡−1x+sin⁡−1x=π2{{\cos }^{-1}}x+{{\sin }^{-1}}x=\dfrac{\pi }{2} and rearrange the equation to get an equation of the form cos⁡−1x=k{{\cos }^{-1}}x=k , where k is constant. Now if k lies in the range of cos⁡−1x{{\cos }^{-1}}x then the equation may have solutions else the equation will never have a solution.

Complete step-by-step answer:
Before starting with the solution to the above question, we will first talk about the required details of different inverse trigonometric ratios. So, we must remember that inverse trigonometric ratios are completely different from trigonometric ratios and have many constraints related to their range and domain.
Now let us draw the graph of cos−1xco{{s}^{-1}}x .

So, looking at the above graphs, we can draw the conclusion that cos⁡−1x{{\cos }^{-1}}x is defined for all real values of x∈[−1,1]x\in [-1,1] , while its range comes out to be [0,π]\left[ 0,\pi \right] . The domain of sin⁡−1x{{\sin }^{-1}}x is same as that of cos⁡−1x{{\cos }^{-1}}x but its range is [−π2,π2]\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] .
Now moving to the solution to the above question, we will start with the simplification of the expression given in the question.
2cos⁡−1x+sin⁡−1x=11π62{{\cos }^{-1}}x+{{\sin }^{-1}}x=\dfrac{11\pi }{6}
We know that cos⁡−1x+sin⁡−1x=π2{{\cos }^{-1}}x+{{\sin }^{-1}}x=\dfrac{\pi }{2} for all x lying in the domain of cos⁡−1x{{\cos }^{-1}}x and sin⁡−1x{{\sin }^{-1}}x .
cos−1x+cos⁡−1x+sin⁡−1x=11π6co{{s}^{-1}}x+{{\cos }^{-1}}x+{{\sin }^{-1}}x=\dfrac{11\pi }{6}
⇒cos−1x+π2=11π6\Rightarrow co{{s}^{-1}}x+\dfrac{\pi }{2}=\dfrac{11\pi }{6}
⇒cos−1x=11π6−π2\Rightarrow co{{s}^{-1}}x=\dfrac{11\pi }{6}-\dfrac{\pi }{2}
⇒cos−1x=11π−3π6=8π6=4π3\Rightarrow co{{s}^{-1}}x=\dfrac{11\pi -3\pi }{6}=\dfrac{8\pi }{6}=\dfrac{4\pi }{3}
Now as 4π3\dfrac{4\pi }{3} is greater than π\pi , we can say that the equation can never be true as cos−1xco{{s}^{-1}}x can never be greater than π\pi , So, the equation has no solution.
Therefore, the answer to the above question is option (a).

Note: Students generally get confused in the range of different inverse trigonometric functions as they very much look the same but are far different. Also, it is important that you check the domain of each inverse trigonometric term in the equation separately before reporting a value of x that satisfies the equation.