Question

What is the number of real solution x of the equation $\displaystyle{\cos^2(x\sin(2x))+\frac{1}{1+x^2} =\cos^2 x+ \sec^2 x}$ ?

Answer 1

Given,

$\sf{\cos^2(x\sin(2x))+\dfrac{1}{1+x^2} =\cos^2 x+ \sec^2 x}$

From this equation I got some features given below.

• For any $\sf{\theta,}$ value of $\sf{\cos^2\theta}$ ranges from 0 to 1.

Hence the term $\sf{\cos^2(x\sin(2x))}$ in the LHS can only have the values from 0 to 1.

• $\sf{0\ \textless\ \dfrac {1}{1+x^2}\leq1\ \forall x\in\mathbb{R}.}$

Therefore the entire LHS can give a sum ranging from 0 to 2. But unfortunately,

• The minimum possible value of the RHS $\sf{\cos^2x+\sec^2x}$ is 2 (Recall the AM ≥ GM inequality).

Hence the LHS should also have a minimum value of 2 but the LHS has 2 as the maximum value. This implies there may exist some solutions for which both the sides of the equation are equalling to 2.

For the RHS, I mean, for the equation $\sf{\cos^2x+\sec^2x=2,}$ has a general solution $\sf{x=n\pi,\ n\in\mathbb{Z}.}$

So taking $\sf{x=n\pi}$ for the LHS equalling to 2, and,

$\longrightarrow\sf{\cos^2(n\pi\sin (2n\pi))+\dfrac {1}{1+(n\pi)^2}=2}$

Since $\sf{\sin (n\pi)=0,}$

$\longrightarrow\sf{1+\dfrac {1}{1+n^2\pi^2}=2}$

This implies,

$\longrightarrow\sf{n^2\pi^2=0}$

Or,

$\longrightarrow\sf{n=0}$

Hence,

$\longrightarrow\sf{x=0}$

This means there exists a unique solution $\sf{x=0}$ for the equation $\sf{\cos^2(x\sin(2x))+\dfrac{1}{1+x^2} =\cos^2 x+ \sec^2 x.}$

Thus the no. of real solutions for the equation is 1.

Answer 2

Answer:

1+cos(2x)‾‾‾‾‾‾‾‾‾‾‾√=2‾√sin−1(sinx)Period of sinx is 2π and that of cos(2x) is 2π2=πHence overall period is L.C.M. of (π,2π)=2πHence we can find total number of solutions in interval of 2π and multiply answer by total intervals of 2π in given interval.Given interval is [−10π,10π][−10π,10π] contains 10π−(−10π)2π=20π2π=10 intervals of 2π1+cos(2x)‾‾‾‾‾‾‾‾‾‾‾√=2‾√sin−1(sinx)⇒1+2cos2x−1‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=2‾√sin−1(sinx)⇒2cos2x‾‾‾‾‾‾‾√=2‾√sin−1(sinx)⇒2‾√∣∣cosx∣∣=2‾√sin−1(sinx)⇒∣∣cosx∣∣=sin−1(sinx)When x∈[−π2,π2]cosx is positive, hence ∣∣cosx∣∣=cosxAnd we know that sin−1(sinx)=x when x∈[−π2,π2]⇒cosx=xDraw the graphs of cosx and x on same graph paper in specified interval.