Question

Find the general solution of \[\sin 3\theta +\cos 2\theta =0\].

Answer 1

Hint: First reduce the equation into a single trigonometric ratio by using the identity that is \[\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta \]. Hence use formula for general solutions which is \[\sin x=\sin \theta \], then \[x=n\pi \pm {{\left( -1 \right)}^{n}}\theta \], where \[\theta \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\] and then find different values by taking n as odd or even.

Complete step-by-step answer:

In the question we are given an equation which is \[\sin 3\theta +\cos 2\theta =0\] and we have to find general solutions or values of \[\theta \] for which equation satisfies.

So, we are given as,

\[\sin 3\theta +\cos 2\theta =0\]

We can rewrite equation as,

\[\cos 2\theta =-\sin 3\theta \]

We know that, \[\sin \left( -x \right)=-\sin x\], so we can use this form as, \[-\sin x=\sin \left( -x \right)\], where \[x=3\theta \] so we get,

\[\cos 2\theta =\sin \left( -3\theta \right)\]

Now, as we know the identity, \[\sin \left( 90-x \right)=\cos x\]. So, we will use it as, \[\cos x=\sin \left( 90-x \right)\], where \[x=2\theta \] so we get,

\[\sin \left( 90-2\theta \right)=\sin \left( -3\theta \right)\]

Now, we will apply formula of general solutions which is if \[\sin x=\sin \alpha \], then \[x=n\pi +{{\left( -1 \right)}^{n}}\alpha \], where \[\alpha \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\]. So, we write equation as,

\[90-2x=n\pi +{{\left( -1 \right)}^{n}}\left( -3x \right)\]

Now, we will take two cases of n one in which n is odd and another one in which n is even.
So, let’s take count for the first case which is odd by putting, \[n=2m+1\].

\[\dfrac{\pi }{2}-2x=n\pi +{{\left( -1 \right)}^{n}}\left( -3x \right)\], can be written as,
\[\dfrac{\pi }{2}-2x=\left( 2m+1 \right)\pi -\left( -3x \right)\]

Or

\[\dfrac{\pi }{2}-2x=\left( 2m+1 \right)\pi +3x\]

So, \[5x=\dfrac{\pi }{2}-\left( 2m+1 \right)\pi \]

Hence, \[x=\dfrac{\pi }{5}\left( \dfrac{1}{2}-\left( 2m+1 \right) \right)\]

\[x=\dfrac{\pi }{5}\left( -2m-\dfrac{1}{2} \right)\]

So, x is equal to \[\dfrac{\pi }{5}\left( -2m-\dfrac{1}{2} \right)\] for m to be any integer.

Now let’s go for \[{{2}^{nd}}\] case which is n is even by putting, \[n=2m\],

\[\dfrac{\pi }{2}-2x=n\pi +{{\left( -1 \right)}^{n}}\left( -3x \right)\], can be written as,

\[\dfrac{\pi }{2}-2x=2m\pi +{{\left( -1 \right)}^{2m}}\left( -3x \right)\]

Or, \[\dfrac{\pi }{2}-2x=2m\pi -3x\]

So, \[3x-2x=\pi \left( 2m-\dfrac{1}{2} \right)\]

Hence, \[x=\pi \left( 2m-\dfrac{1}{2} \right)\]

So, x is equal to \[\pi \left( 2m-\dfrac{1}{2} \right)\] for any integer.

Hence, the general solutions are \[\dfrac{\pi }{5}\left( -2m-\dfrac{1}{2} \right)\] and \[\pi \left( 2m-\dfrac{1}{2} \right)\].

Note: Generally in the questions students confuse like after changing value according to standard angles how to write general solutions so we use here formula; if \[\sin x=\sin \theta \], then \[x=n\pi +{{\left( -1 \right)}^{n}}\theta \], where \[\theta \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\].