Question

How do you find the general solution for \[\sin x = \dfrac{{ - \sqrt 2 }}{2}\]?

Answer 1

Hint: Solution of a trigonometric equation is the value of the unknown angle that satisfies the equation. A solution generalised by the means of periodicity is known as the general solution. To find the general solution of the above trigonometric equation we will first solve the RHS. Then we will express it in the form of sine of an angle, then we will use the formula for the general solution of \[\sin \theta = \sin \alpha \].

Formula used:
The general solution for \[\sin \theta = \sin \alpha \], is \[\theta = n\pi + {\left( { - 1} \right)^n}\alpha ,n \in z\].

Complete step by step answer:
We have the equation as;
\[\sin x = \dfrac{{ - \sqrt 2 }}{2}\]
We can see that in the RHS we can cancel \[\sqrt 2 \] from the numerator and denominator. So, simplifying the RHS we get;
\[ \Rightarrow \sin x = - \dfrac{1}{{\sqrt 2 }}\]
Now we know that \[\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]. So, we will replace the RHS using this. So, we get;
\[ \Rightarrow \sin x = - \sin \dfrac{\pi }{4}\]
Also, we know that \[ - \sin \theta = \sin \left( { - \theta } \right)\]. So, we will again use this property of sine function to replace the RHS. So, we get;
\[ \Rightarrow \sin x = \sin \left( {\dfrac{{ - \pi }}{4}} \right)\]
Now this is in the form \[\sin \theta = \sin \alpha \]. So, we will now use the general solution for this type of equation. We know;
\[\theta = n\pi + {\left( { - 1} \right)^n}\alpha ,n \in z\]
So, we get;
\[ \Rightarrow x = n\pi + {\left( { - 1} \right)^n}\left( {\dfrac{{ - \pi }}{4}} \right)\]
We can also write it as;
\[ \Rightarrow x = n\pi - {\left( { - 1} \right)^n}\dfrac{\pi }{4}\]

Note:
One thing to note here is that if in any other question we get a condition like \[\csc \theta = \csc\alpha \], then also we can use the same solution rule as we have used for \[\sin \theta = \sin \alpha \]. This is because when \[\csc \theta = \csc \alpha \], then we indirectly have \[\sin \theta = \sin \alpha \]. Similarly, we can use the solution rule for \[\cos \theta = \cos \alpha \], for writing the solution for \[\sec \theta = \sec \alpha \]. Similarly, from the tangent function we can write for the cotangent function.