Question

How do you solve $\sin 2x\cos x+\cos 2x\sin x=\dfrac{\sqrt{2}}{2}$?

Answer 1

Consider the L.H.S. of the given trigonometric equation and use the sum of angle formula given by: - $\sin a\cos b+\cos a\sin b=\sin \left( a+b \right)$ to simplify. Now, simplify the R.H.S. by cancelling the common factor. Use the general solution formula of sine function given as: - if $\sin a=\sin b$ then $a=n\pi +{{\left( -1 \right)}^{n}}b$, where $n\in$ integers, to get the answer.

Complete step by step answer:
Here, we have been provided with the trigonometric equation: - $\sin 2x\cos x+\cos 2x\sin x=\dfrac{\sqrt{2}}{2}$ and we are asked to solve it.
That means we have to find the value of x.
$\because \sin 2x\cos x+\cos 2x\sin x=\dfrac{\sqrt{2}}{2}$
As we can see that the L.H.S. of the above equation is of the form: - $\sin a\cos b+\cos a\sin b$ whose simplified form is given as: - $\sin \left( a+b \right)$. So, using this conversion formula in the L.H.S., we get,

& \Rightarrow \sin \left( 2x+x \right)=\dfrac{\sqrt{2}}{2} \\\ & \Rightarrow \sin 3x=\dfrac{\sqrt{2}}{2} \\\ \end{aligned}$$ The above equation can be written as: - $$\Rightarrow \sin 3x=\dfrac{\sqrt{2}}{{{\left( \sqrt{2} \right)}^{2}}}$$ Cancelling the common factors in the R.H.S., we get, $$\Rightarrow \sin 3x=\dfrac{1}{\sqrt{2}}$$ Now, we know that $$\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$$, so replacing $$\dfrac{1}{\sqrt{2}}$$ with $$\sin \dfrac{\pi }{4}$$ in the above relation, we get, $$\Rightarrow \sin 3x=\sin \left( \dfrac{\pi }{4} \right)$$ The above equation now is of the form $$\sin a=\sin b$$ whose general solution is given by the relation: - $$a=n\pi +{{\left( -1 \right)}^{n}}b$$, where $$n\in $$ integers. So, using the general solution formula, we get, $$\Rightarrow \sin 3x=n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{\pi }{4} \right),n\in z$$ Dividing both the sides with 3, we get, $$\begin{aligned} & \Rightarrow x=\dfrac{n\pi }{3}+{{\left( -1 \right)}^{n}}\dfrac{\pi }{4\times 3},n\in z \\\ & \Rightarrow x=\dfrac{n\pi }{3}+{{\left( -1 \right)}^{n}}\dfrac{\pi }{12},n\in z \\\ & \Rightarrow x=\dfrac{\pi }{12}\left[ 4n+{{\left( -1 \right)}^{n}} \right],n\in z \\\ \end{aligned}$$ Hence, $$x=\dfrac{\pi }{12}\left[ 4n+{{\left( -1 \right)}^{n}} \right]$$ is the general solution of the given trigonometric equation. **Note:** One may note that here we have found the general solution and not the principal solution of the given equation. This is because we were not provided with any particular range of angle between which we were required to find the value of x. If any range of angle would have been provided then we had to find the values of x between that range by substituting suitable values of n. You must remember all the formulas like: - $$\sin \left( a\pm b \right),\cos \left( a\pm b \right),\tan \left( a\pm b \right)$$ etc. Always remember that the value of ‘n’ will be an integer.