Question

Solve in terms of x: $2\sin x = - 1,\;0 \leqslant x \leqslant 4\pi $

Answer 1

Hint: Often we will solve a trigonometric equation over a specified interval as mentioned in this problem. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period.
Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity. Substitute the trigonometric expression with a single variable if required. Solve the equation the same way an algebraic equation would be solved. Substitute the trigonometric expression back in for the variable in the resulting expressions.

Complete step-by-step answer:
$2\sin x = - 1,\;0 \leqslant x \leqslant 4\pi $
Divide each term by 2 and simplify
$\dfrac{{2\sin (x)}}{2} = \dfrac{{ - 1}}{2}$
Cancel the common factor of 2
$ \Rightarrow \sin (x) = \dfrac{{ - 1}}{2}$
Move the negative in front of the fraction.
$ \Rightarrow \sin (x) = - \dfrac{1}{2}$
Now lets find the value of x when $\sin (x) = - \dfrac{1}{2}$
$ \Rightarrow x = - \dfrac{\pi }{6}$
The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2\pi $, to find a reference angle. Next add this reference angle to $\pi $to find the solution in the third quadrant.
$ \Rightarrow x = 2\pi + \dfrac{\pi }{6} + \pi $
Simplify the expression to find the second solution.
$ \Rightarrow x = 2\pi + \dfrac{\pi }{6} + \pi = \dfrac{{19\pi }}{6}$
To find the second solution, subtract the solution from $2\pi $, to find a reference angle
$ \Rightarrow x = \dfrac{{19\pi }}{6} - 2\pi = \dfrac{{7\pi }}{6}$
Add $2\pi $ to every negative angle to get positive angles.
$ \Rightarrow x = - \dfrac{\pi }{6} + 2\pi = \dfrac{{11\pi }}{6}$
The period of the \[sin\left( x \right)\] function is \[2\pi \]so values will repeat every \[2\pi \]radians in both directions.
\[x = \dfrac{{7\pi }}{6} + 2\pi n,\dfrac{{11\pi }}{6} + 2\pi n\], for any integer n.
If n= 0, then value of $x = \dfrac{{7\pi }}{6},\dfrac{{11\pi }}{6}$
If n= 1, then value of $x = \dfrac{{19\pi }}{6},\dfrac{{23\pi }}{6}$
These are the two values which are applicable as above these x values will be increased from $4\pi $.

So the four values of x for equation $2\sin x = - 1,\;0 \leqslant x \leqslant 4\pi $ are $\left( {\dfrac{{7\pi }}{6},\dfrac{{11\pi }}{6},\dfrac{{19\pi }}{6},\dfrac{{23\pi }}{6}} \right)$.

Note: The domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is $2\pi $. In other words, every $2\pi $ units, the y-values repeat. If we need to find all possible solutions, then we must add $2\pi k$ where k is an integer, to the initial solution.