Question

How do you find the general solution for \[\sin x = - \dfrac{1}{2}\]?

Answer 1

Hint: Use the concept of general solution of a trigonometric function. Write the value of the right hand side of the equation in terms of sine of an angle. Use a quadrant diagram to write the value for the right side of the equation.
* We know the values of all trigonometric angles are positive in the first quadrant.
Values of only \[\sin \theta \] are positive in the second quadrant.
Values of only \[\tan \theta \] are positive in the third quadrant.
Values of only \[\cos \theta \] are positive in the fourth quadrant.

Complete step-by-step solution:
We have to find the general solution of \[\sin x = - \dfrac{1}{2}\]
We will write the right hand side of the equation in terms of sine of an angle.
We know that \[\sin \dfrac{\pi }{6} = \dfrac{1}{2}\]
Substitute the value of \[\sin \dfrac{\pi }{6} = \dfrac{1}{2}\] in right hand side of the equation
\[ \Rightarrow \sin x = - \left( {\sin \dfrac{\pi }{6}} \right)\].................… (1)
Now we know from the quadrant diagram that sine of an angle is negative in only the third and fourth quadrant. Also we know that sine is an odd function i.e. \[\sin ( - x) = - \sin x\].
So we can write the value of angle inside the sine function as \[\sin ( - x) = \sin (\pi + x)\]or\[\sin ( - x) = \sin (2\pi - x)\].
Then the equation (1) becomes
\[ \Rightarrow \sin x = \sin \left( {\pi + \dfrac{\pi }{6}} \right)\] or \[\sin x = \sin \left( {2\pi - \dfrac{\pi }{6}} \right)\]
Since both sides have same function, we can take inverse sine function on both sides of the equation
\[ \Rightarrow {\sin ^{ - 1}}\left\{ {\sin x} \right\} = {\sin ^{ - 1}}\left\{ {\sin \left( {\pi + \dfrac{\pi }{6}} \right)} \right\}\] or \[{\sin ^{ - 1}}\left\{ {\sin x} \right\} = {\sin ^{ - 1}}\left\{ {\sin \left( {2\pi - \dfrac{\pi }{6}} \right)} \right\}\]
Cancel inverse function by the function
\[ \Rightarrow x = \left( {\pi + \dfrac{\pi }{6}} \right)\] or \[x = \left( {2\pi - \dfrac{\pi }{6}} \right)\]
Take LCM on both terms
\[ \Rightarrow x = \left( {\dfrac{{6\pi + \pi }}{6}} \right)\] or \[x = \left( {\dfrac{{12\pi - \pi }}{6}} \right)\]
\[ \Rightarrow x = \left( {\dfrac{{7\pi }}{6}} \right)\] or \[x = \left( {\dfrac{{11\pi }}{6}} \right)\]
We know that general solution can be written as \[2n\pi + x\], where n is an integer
So, the general solution is \[2n\pi + \dfrac{{7\pi }}{6}\] or \[2n\pi + \dfrac{{11\pi }}{6}\]

\[\therefore \]General solution for \[\sin x = - \dfrac{1}{2}\] is \[2n\pi + \dfrac{{7\pi }}{6}\] or \[2n\pi + \dfrac{{11\pi }}{6}\].

Note: Many students make the mistake of writing the value of the sine function in the first step as they don’t remember the values. Keep in mind we can use table for trigonometric terms if we don’t remember the values at some common angles \[{0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ },{90^ \circ }\]

ANGLEFUNCTION	\[{0^ \circ }\]	\[{30^ \circ }\]	\[{45^ \circ }\]	\[{60^ \circ }\]	\[{90^ \circ }\]
Sin	0	\[\dfrac{1}{2}\]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{{\sqrt 3 }}{2}\]	1
Cos	1	\[\dfrac{{\sqrt 3 }}{2}\]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{1}{2}\]	0
Tan	0	\[\dfrac{1}{{\sqrt 3 }}\]	1	\[\sqrt 3 \]	Not defined