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

Hint:
Here, first form a quadratic equation using the formula$\cos 2\theta = 1 - 2{\sin ^2}\theta $. Once the quadratic equation is formed compare it with standard quadratic equation which is given as $a{x^2} + bx + c = 0$.Then use the formula of discriminant which is given as-
$ \Rightarrow D = {b^2} - 4ac$ to find if the equation will have real roots or not.

Complete step by step solution:
Given $\sin 2x + \cos 4x = 2$
We have to find the number of solutions of this function in the interval$\left( {0,2\pi } \right)$.
We can find the solution by changing the given equation into a quadratic equation and by finding its discriminant.
We know that $\cos 2\theta = 1 - 2{\sin ^2}\theta $
On applying this formula in the function where $\theta = 2x$ we get,
$ \Rightarrow \sin 2x + 1 - 2{\sin ^2}2x = 2$
On rearranging the equation we get,
$ \Rightarrow \sin 2x - 2{\sin ^2}2x + 1 = 2$
Now on transferring one to the right side of the equation we get,
$ \Rightarrow \sin 2x - 2{\sin ^2}2x = 2 - 1$
On solving we get,
$ \Rightarrow \sin 2x - 2{\sin ^2}2x = 1$
On again transferring one to left side we get,
$ \Rightarrow \sin 2x - 2{\sin ^2}2x - 1 = 0$
On multiplying the equation with negative sign and rearranging again we get,
$ \Rightarrow 2{\sin ^2}2x - \sin 2x + 1 = 0$
Now this equation is in the form of quadratic equation in $\sin 2x$
On comparing it with the standard quadratic equation $a{x^2} + bx + c = 0$ we get,
$ \Rightarrow a = 2,b = - 1,c = 1$
Now we know that the formula of discriminant is given as-
$ \Rightarrow D = {b^2} - 4ac$
On putting the given values in the formula we get,
$ \Rightarrow D = {\left( { - 1} \right)^2} - 4 \times 2 \times 1$
On solving we get,
$ \Rightarrow D = 1 - 8$
On subtraction we get,
$ \Rightarrow D = - 7$
Here,
$ \Rightarrow D = - 7 < 0$
And we know that if the value of discriminant is less than zero then the equation does not have any real roots.
Hence there are no real solutions of this equation so we can say that the number of solutions for the given equation is zero.

Hence the correct answer is A.

Note:
Here the student may go wrong if the use the formula of $\sin 2\theta = 2\sin \theta \cos \theta $ because then the equation will become-
$ \Rightarrow $ $2\sin x\cos x + \cos 4x = 2$
 Here no quadratic equation can be formed and the given equation will become more complex. This will make the equation difficult to solve hence we use the formula $\cos 2\theta = 1 - 2{\sin ^2}\theta $ to form a quadratic equation.