Question

The number of solution(s) of $\sin 2x + \cos 4x = 2$in the interval $\left( {0,2\pi } \right)$is
A. 0
B. 2
C. 3
D. 4

Answer 1

Firstly, use the formula of $\cos 2x = 1 - 2{\sin ^2}x$ in order to make the equation with all the same terms and hence finally we will be left with the quadratic equation and so it’s discriminant can be calculated and idea about its nature and number of roots can be calculated.

Complete step by step answer:

As the given equation is $\sin 2x + \cos 4x = 2$
So, using the formula of $\cos 4x = 1 - 2{\sin ^2}2x$ converts the value of the equation into a single format.
So, it can be given as
\Rightarrow $$$$\sin 2x + 1 - 2{\sin ^2}2x = 2
On rearranging we get,
\Rightarrow $$$$2{\sin ^2}2x - \sin 2x + 1 = 0
Now, we can see that the quadratic equation is obtained and so let $\sin x = t$and place it in above equation hence the equation be $2{t^2} - t + 1 = 0$.
Now calculating the discriminant of the quadratic equation in order to obtain the idea of the number of roots of the equation.
As $D = {b^2} - 4ac$
So on substituting the values, we get,
\Rightarrow $$$$D = {\left( { - 1} \right)^2} - 4\left( 2 \right)\left( 1 \right)
On calculating,
\Rightarrow $$$$D = 1 - 8
And on subtracting ,
\Rightarrow $$$$D = - 7
As the discriminant is less than zero,
Hence, no real roots for the equation and hence no solution exists for the equation $\sin 2x + \cos 4x = 2$ in the interval $\left( {0,2\pi } \right)$
Hence, option (A) is correct answer.

Note: In algebra, a quadratic equation is an equation that can be rearranged in standard form as where x represents an unknown, and a, b, and c represent known numbers, where $a \ne 0$. If $a \ne 0$, then the equation is linear, not quadratic, as there is no term.
The nature of roots can be obtained using the value of $D = {b^2} - 4ac$
If D is greater than 0, then the roots are real and distinct.
If D is less than 0, then the roots are imaginary, or we can say no real roots.
If D is 0 then we have real and equal roots.