Question

If \[3\cos x\ne -2\sin x\], then the general solution of \[{{\sin }^{2}}x-\cos 2x=2-\sin 2x\] is
(A) \[n\pi +\dfrac{\pi }{2},n\in Z\]
(B) \[\dfrac{n\pi }{2},n\in Z\]
(C) \[4n\pm 1\dfrac{\pi }{2},n\in Z\]
(D) \[2n-\pi ,n\in Z\]

Answer 1

Hint: We are given an expression involving trigonometric functions and using that we have to find the general solution of the given expression. We will expand the expression using the identities, \[\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x\] and \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]. Then, we will cancel out the similar terms on either side of the equality. We will then have an expression as, \[\cos x\left( 2\sin x-3\cos x \right)=0\]. But, we are given that, \[3\cos x\ne -2\sin x\], so, we will get, \[\cos x=0\]. And accordingly we will have the solution of ‘x’. Hence, we will have the required general solution.

Complete step by step answer:
According to the given question, we are given a mathematical statement, which is, \[3\cos x\ne -2\sin x\] and using this we are asked to find the general solution of \[{{\sin }^{2}}x-\cos 2x=2-\sin 2x\].
The expression we have is,
\[{{\sin }^{2}}x-\cos 2x=2-\sin 2x\]
We will expand using the identity of \[\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x\], we get,
\[\Rightarrow {{\sin }^{2}}x-\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)=2-\sin 2x\]
Opening up the brackets, we get the expression as,
\[\Rightarrow {{\sin }^{2}}x-{{\cos }^{2}}x+{{\sin }^{2}}x=2-\sin 2x\]
\[\Rightarrow 2{{\sin }^{2}}x-{{\cos }^{2}}x=2-\sin 2x\]
We know that, \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\], we can write it as, \[{{\sin }^{2}}x=1-{{\cos }^{2}}x\], so we have,
\[\Rightarrow 2\left( 1-{{\cos }^{2}}x \right)-{{\cos }^{2}}x=2-\sin 2x\]
We get the expression as,
\[\Rightarrow 2-2{{\cos }^{2}}x-{{\cos }^{2}}x=2-\sin 2x\]
\[\Rightarrow -3{{\cos }^{2}}x=-\sin 2x\]
We know that, \[\sin 2x=2\sin x\cos x\], so, we have,
\[\Rightarrow -3{{\cos }^{2}}x=-2\sin x\cos x\]
We will now take out the common terms and we get the expression as,
\[\Rightarrow \cos x\left( 2\sin x-3\cos x \right)=0\]
We are given that, \[3\cos x\ne -2\sin x\], so \[2\sin x-3\cos x=0\] is not possible, so we have,
\[\cos x=0\]
Therefore, we have \[x=n\pi +\dfrac{\pi }{2},n\in Z\].

So, the correct answer is “Option A”.

Note: The expression is usually solved for its solution by making the expression look similar to the given statements like in the above solution, we made the term in the expression similar to \[3\cos x\ne -2\sin x\]. This way we are on the correct track to find the correct solution of ‘x’. Also, use the correct identities while simplifying the expression. And cosine function gives a value 0 for right angles and multiples of 90 degrees.