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Common roots of the equations\[2{\sin ^2}x + {\sin ^2}2x = 2\]and \[\sin 2x + \cos 2x = \tan x\]are
A. \[x = (2n - 1)\frac{\pi }{2}\]
В. \[x = (2n + 1)\frac{\pi }{4}\]
C. \[x = (2n + 1)\frac{\pi }{3}\]
D. None of these




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Last updated date: 19th Jul 2024
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Answer
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Hint: When algebraic procedures are used to solve a trigonometric problem, the term "root loss" describes how some roots are lost. To precisely determine the answer, it is necessary to perform the superfluous root check. In this case, the common roots for the equation\[2{\sin ^2}x + {\sin ^2}2x = 2\]and \[\sin 2x + \cos 2x = \tan x\]is found by using trigonometry identities





Complete step by step solution:We have given the equation, as per the question:
\[2{\sin ^2}x + {\sin ^2}2x = 2\]-- (i)
\[\sin 2x + \cos 2x = \tan x\]-- (ii)
First we can solve equation (i)
\[{\sin ^2}2x = 2{\cos ^2}x\]
\[ \Rightarrow {\rm{ }}4{\sin ^2}x{\cos ^2}x = 2{\cos ^2}x\]
In order to simplify further, we have to take the common terms outside:
\[ \Rightarrow {\cos ^2}x\left( {2{{\sin }^2}x - 1} \right) = 0\]
Solve the terms in the parentheses, so as to make the expression less complicated to evaluate:
\[ \Rightarrow 2{\cos ^2}x\cos 2x = 0\]
Simplify in terms of cosine:
\[ \Rightarrow \cos x = 0\]Or\[\cos 2x = 0\]
Also can be written as:
\[x = (2n + 1)\frac{\pi }{2}\]Or\[x = (2n + 1)\frac{\pi }{4}\],\[n \in Z\]--(iii)
Now, we will solve Eq. (ii)
\[\dfrac{{2\tan x + 1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}} = \tan x\]
Solve and simplify the denominator to make the fraction to expression:
\[ \Rightarrow \tan 3x + {\tan ^2}x - \tan x - 1 = 0\]
Factor the equation for less complicate to solve:
\[ \Rightarrow (\tan 2x - 1)(\tan x + 1) = 0\]
Hence after solving,
\[ \Rightarrow \tan x = \pm 1 \Rightarrow x = n\pi \pm \frac{\pi }{4},n \in Z\]-- (iv)
From equations (iii) and (iv);
Hence the common roots are \[(2n + 1)\frac{\pi }{4}\]



Option ‘B’ is correct

Note: The identities of triangles are frequently forgotten by students. Failing to adjust or change the scope of. Getting the right side of the equation to equal zero is the first step in solving the problem. We factor next. After that, we solve by setting each element to zero. This kind of problem can be solved in a number of ways. There may be different formulae to handle this kind of problem as well, but we must think carefully and select the best approach and formula. Trigonometry exercise and formula study will help us achieve this.