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Question: Solve the equation for \(x\): \({\sin ^{ - 1}}x + {\sin ^{ - 1}}(1 - x) = {\cos ^{ - 1}}x,x \ne 0\...

Solve the equation for xx:
sin⁡−1x+sin⁡−1(1−x)=cos⁡−1x,x≠0{\sin ^{ - 1}}x + {\sin ^{ - 1}}(1 - x) = {\cos ^{ - 1}}x,x \ne 0

Explanation

Solution

Hint: Substitute for xx as a straight trigonometric term, like sin⁡y\sin y and then reduce the number of inverse trigonometric terms by rearranging the terms and using trigonometric relations.

Complete step-by-step answer:
Our problem is as follows:
sin⁡−1x+sin⁡−1(1−x)=cos⁡−1x{\sin ^{ - 1}}x + {\sin ^{ - 1}}(1 - x) = {\cos ^{ - 1}}x
To solve for xx, will have to reduce the number of inverse trigonometric terms first. To do this generally, we will introduce another variable yy such that:
x=sin⁡yx = \sin y. Therefore,
(1−x)=(1−sin⁡y); x=cos⁡(π2−y) \begin{gathered} (1 - x) = (1 - \sin y); \\\ x = \cos (\dfrac{\pi }{2} - y) \\\ \end{gathered}
We will substitute these assumptions in our problem and see whether we can simplify our problem. Generally, the assumptions are considered in such a way that maximum number of inverse terms are reduced to simple linear terms in a few number of steps. There is no standard methodology to determine these assumptions and you will gradually develop this instinct with practice of more similar problems. So moving ahead with our problem, we get,
⇒sin⁡−1(sin⁡y)+sin⁡−1(1−sin⁡y)=cos⁡−1[cos⁡(π2−y)] ⇒y+sin⁡−1(1−sin⁡y)=π2−y ⇒sin⁡−1(1−sin⁡y)=π2−2y ⇒1−sin⁡y=sin⁡(π2−2y) ⇒cos⁡2y+sin⁡y−1=0  \Rightarrow {\sin ^{ - 1}}(\sin y) + {\sin ^{ - 1}}(1 - \sin y) = {\cos ^{ - 1}}\left[ {\cos \left( {\dfrac{\pi }{2} - y} \right)} \right] \\\ \Rightarrow y + {\sin ^{ - 1}}(1 - \sin y) = \dfrac{\pi }{2} - y \\\ \Rightarrow {\sin ^{ - 1}}(1 - \sin y) = \dfrac{\pi }{2} - 2y \\\ \Rightarrow 1 - \sin y = \sin \left( {\dfrac{\pi }{2} - 2y} \right) \\\ \Rightarrow \cos 2y + \sin y - 1 = 0 \\\
Now we know that, cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2{\sin ^2}\theta . Using this identity, we get,
⇒1−2sin⁡2y+sin⁡y−1=0 ⇒sin⁡y(1−2sin⁡y)=0  \Rightarrow 1 - 2{\sin ^2}y + \sin y - 1 = 0 \\\ \Rightarrow \sin y(1 - 2\sin y) = 0 \\\
We will two values for sin⁡y\sin y, which are as follows:
sin⁡y=0\sin y = 0 and sin⁡y=12\sin y = \dfrac{1}{2}. Now we will take back this to our required variable, xx.
Therefore, we have two solutions, x=0x = 0 or x=12x = \dfrac{1}{2}.

Note: You will get the same solution for any assumption you do in the beginning of the problem but it will lead to more steps in between the problem and the solution.