Question

# The number of solution of $\sin 3x=\cos 2x$ in the interval $\left( \dfrac{\pi }{2},\pi \right)$ is:(a) 1(b) 2(c) 3(d) 4

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Hint: We know that according to trigonometric identities $\sin 3x$ is equal to $3\sin x-4{{\sin }^{3}}x$ and also $\cos 2x$ is equal to $1-2{{\sin }^{2}}x$ so substituting these values of $\sin 3x\And \cos 2x$ in the above equation and then we will get the cubic equation in $\sin x$ and then solve this cubic equation and find the roots of this equation and hence, find the value of x.

The equation given in the above problem is:
$\sin 3x=\cos 2x$………… Eq. (1)
We know from the trigonometric identities that:
\begin{align} & \sin 3x=3\sin x-4{{\sin }^{3}}x \\ & \cos 2x=1-2{{\sin }^{2}}x \\ \end{align}
Using the above relations in eq. (1) we get,
$3\sin x-4{{\sin }^{3}}x=1-2{{\sin }^{2}}x$
Arranging the above equation we get,
$4{{\sin }^{3}}x-2{{\sin }^{2}}x-3\sin x+1=0$
Substituting $\sin x=t$ in the above equation we get,
$4{{\left( t \right)}^{3}}-2{{t}^{2}}-3t+1=0$
If we put $t=1$ in the above equation we get,
\begin{align} & 4-2-3+1=0 \\ & \Rightarrow 5-5=0 \\ & \Rightarrow 0=0 \\ \end{align}
L.H.S = R.H.S in the above equation hence $t=1$ is satisfying the above equation.
Now, on dividing $4{{\left( t \right)}^{3}}-2{{t}^{2}}-3t+1$ by $\left( t-1 \right)$ we get,
t-1\overset{4{{t}^{2}}+2t-1}{\overline{\left){\begin{align} & 4{{t}^{3}}-2{{t}^{2}}-3t+1 \\ & \dfrac{4{{t}^{3}}-4{{t}^{2}}}{\begin{align} & 0+2{{t}^{2}}-3t \\ & 0+\dfrac{2{{t}^{2}}-2t}{\begin{align} & -t+1 \\ & \dfrac{-t+1}{00} \\ \end{align}} \\ \end{align}} \\ \end{align}}\right.}}
Now, we are going to find the roots of this equation $4{{t}^{2}}+2t-1=0$ by Shree Dharacharya rule as follows:
We are solving the roots of the equation by Shree Dharacharya formula.
Discriminant of a quadratic equation is denoted by D.
For the quadratic equation$a{{t}^{2}}+bt+c=0$the value of D is:
$D={{b}^{2}}-4ac$
Comparing this value of D with the given quadratic equation$4{{t}^{2}}+2t-1=0$we get,
\begin{align} & D={{\left( 2 \right)}^{2}}-4\left( 4 \right)\left( -1 \right) \\ & \Rightarrow D=4+16 \\ & \Rightarrow D=20 \\ \end{align}
The discriminant formula for finding the roots of quadratic equation$a{{t}^{2}}+bt+c=0$is:
$t=\dfrac{-b\pm \sqrt{D}}{2a}$
Comparing the above value of t with the given quadratic equation $4{{t}^{2}}+2t-1=0$ we get,
\begin{align} & t=\dfrac{-2\pm \sqrt{20}}{8} \\ & \Rightarrow t=\dfrac{-2\pm 2\sqrt{5}}{8} \\ \end{align}
Dividing numerator and denominator by 2 we get,
$t=\dfrac{-1\pm \sqrt{5}}{4}$
Hence, the solutions of “t” that we are getting are:
$t=1$;
$t=\dfrac{-1\pm \sqrt{5}}{4}$
Now, we have assumed that $t=\sin x$ so substituting $t=\sin x$ in the above equation we get,
$\sin x=1$
$\sin x=\dfrac{-1\pm \sqrt{5}}{4}$
Now, we have to find the solutions which lie in the interval $\left( \dfrac{\pi }{2},\pi \right)$ so solving the above equations in x we get,
$\sin x=1$
The above equation is true when $x=\dfrac{\pi }{2}$ and the solution should lie in the interval $\left( \dfrac{\pi }{2},\pi \right)$. But the solution that we are getting includes the boundary values of this interval but it is given that we should not include boundary values. Hence, $x=\dfrac{\pi }{2}$ is not a required solution.
$\sin x=\dfrac{-1\pm \sqrt{5}}{4}$…………. Eq. (2)
When we take negative value in the above equation then $\sin x$ becomes negative and we have to find the solutions in the interval $\left( \dfrac{\pi }{2},\pi \right)$ and in this interval the value of sine is always positive so negative value of the above solution is rejected.
Now, when we take positive value of the expression given in eq. (2) we get,
$\sin x=\dfrac{-1+\sqrt{5}}{4}$
The two values of x are possible where the above equation holds true which is in the first and second quadrant but we are asked the solution which lies in the second quadrant i.e. $\left( \dfrac{\pi }{2},\pi \right)$ so only one solution is possible.
Hence, the correct option is (a).

Note: The point to be taken care of in this problem is that while writing the solutions of the eq. (2), make sure you have considered that the solutions should lie in the interval $\left( \dfrac{\pi }{2},\pi \right)$ because on solving the eq. (2), you will find that there are solutions which does not lie in the interval $\left( \dfrac{\pi }{2},\pi \right)$.
$\sin x=\dfrac{-1\pm \sqrt{5}}{4}$
When we consider minus sign then the value of sine will be negative and if we don’t consider that the solutions should lie in the interval $\left( \dfrac{\pi }{2},\pi \right)$ then you will end up having solutions in which sine is negative and also x lies in the first quadrant.
Generally, in the rush of solving exam papers, students make such mistakes.