Question

How do you solve \[\sin x\cos x+\cos x=0\]?

Answer 1

Hint: Take \[\cos x\] common from the two terms and write the remaining terms inside the bracket. Now, substitute the product of these two terms equal to 0 one – by – one and consider the two cases. Use the formula for general solution of trigonometric equations given as: - if \[\sin a=\sin b\] then \[a=n\pi +{{\left( -1 \right)}^{n}}b\] and if \[\cos k=0\] then \[k=\left( 2m+1 \right)\dfrac{\pi }{2}\], to get the answers. Here, both m and n must be integers.

Complete step by step answer:
Here, we have been provided with the trigonometric equation \[\sin x\cos x+\cos x=0\] and we are asked to solve this equation. That means we have to find the value of x, i.e., general solutions.
\[\because \sin x\cos x+\cos x=0\]
Taking \[\cos x\] common from the two terms, we get,
\[\Rightarrow \cos x\left( \sin x+1 \right)=0\]
Now, we have a product of two terms equal to 0, so either of them can be equal to 0. Substituting each term equal to 0, we get,
\[\Rightarrow \cos x=0\] or \[\left( \sin x+1 \right)=0\]
\[\Rightarrow \cos x=0\] or \[\sin x=-1\]
Considering the two cases one – by – one, we have,
(1) Case (i): - \[\cos x=0\]
We know that the value of the cosine function is equal to 0 when we have odd multiples of \[\dfrac{\pi }{2}\], so mathematically we have,
\[\Rightarrow \cos x=0\]
\[\Rightarrow x=\left( 2m+1 \right)\dfrac{\pi }{2}\], where m \[\in \] integers.
(2) Case (ii): - \[\sin x=-1\]
We can write the above equation as: -
\[\Rightarrow \sin x=-\sin \left( \dfrac{\pi }{2} \right)\]
Now, we know that \[-\sin \theta =\sin \left( -\theta \right)\], so we have,
\[\Rightarrow \sin x=\sin \left( \dfrac{-\pi }{2} \right)\]
We know that the general solution of the equation \[\sin a=\sin b\]is given as: - \[a=n\pi +{{\left( -1 \right)}^{n}}b\], so using this formula for the above relation, we have,
\[\Rightarrow x=n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{-\pi }{2} \right)\], where n \[\in \] integers.
\[\Rightarrow x=n\pi -{{\left( -1 \right)}^{n}}\left( \dfrac{\pi }{2} \right)\], where n \[\in \] integers.

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 m and n. You must remember the formulas for general solutions of all six trigonometric functions. Always remember that the values of ‘m’ and ‘n’ will be integers.