Question

Find the general solution of \[\sin \theta =\tan \theta \].

Answer 1

Hint: First rearrange the given equation and write the equation as \[\sin x=0\] and \[\cos x\] then use the formula for general equation which is if \[\sin \theta =\sin \alpha \], then \[\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha \] where \[\alpha \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\]. If \[\cos \theta =\cos \alpha \] then \[\theta =2n\pi \pm \alpha \] where, \[\alpha \in \left( 0,\pi \right]\].

Complete step-by-step answer:

In the question we are given an equation which is \[\sin \theta =\tan \theta \] and we have to find general solutions or general values of \[\theta \].

So, we are given that,

\[\sin \theta =\tan \theta \]

Now, we will use formula \[\tan \theta \] which is equal to \[\dfrac{\sin \theta }{\cos \theta }\] and rewrite the equation as,

\[\sin \theta =\dfrac{\sin \theta }{\cos \theta }\]

Now, we will rearrange equation as,

\[\sin \theta \cos \theta -\sin \theta =0\]

Now let’s take \[\sin \theta \] common so we get,

\[\sin \theta \left( \cos \theta -1 \right)=0\]

So, we can say that, either \[\sin x=0\] or \[\cos x=1\].

Now, we will take the first case which is \[\sin x=0\].

So, we can write \[\sin x=0\] as \[\sin x=\sin 0\]. Hence, x is equal to \[n\pi +{{\left( -1

\right)}^{n}}\left( 0 \right)\] or \[\left( n\pi \right)\], where n is an integer.

Now, we will go for \[{{2}^{nd}}\] case which is \[\cos x=1\].

So, we can write \[\cos x=1\] as \[\cos x=\cos 0\]. Hence, x is equal to \[2n\pi +0\] or \[\left( 2m\pi \right)\], where m is an integer.

Hence, the general or required solutions are \[x=2m\pi \] or \[x=n\pi \].

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 \[\theta \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\]; if \[\cos x=\cos \theta \], then \[x=2n\pi \pm \theta \], where \[\theta \in \left( 0,\pi \right]\].