Solveeit Logo
Login

Question

Question: Solve in terms of x: \(2\sin x = - 1,\;0 \leqslant x \leqslant 4\pi \)...

Solve in terms of x: 2sin⁡x=−1,  0⩽x⩽4π2\sin x = - 1,\;0 \leqslant x \leqslant 4\pi

Explanation

Solution

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:
2sin⁡x=−1,  0⩽x⩽4π2\sin x = - 1,\;0 \leqslant x \leqslant 4\pi
Divide each term by 2 and simplify
2sin⁡(x)2=−12\dfrac{{2\sin (x)}}{2} = \dfrac{{ - 1}}{2}
Cancel the common factor of 2
⇒sin⁡(x)=−12\Rightarrow \sin (x) = \dfrac{{ - 1}}{2}
Move the negative in front of the fraction.
⇒sin⁡(x)=−12\Rightarrow \sin (x) = - \dfrac{1}{2}
Now lets find the value of x when sin⁡(x)=−12\sin (x) = - \dfrac{1}{2}
⇒x=−π6\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 2π2\pi , to find a reference angle. Next add this reference angle to π\pi to find the solution in the third quadrant.
⇒x=2π+π6+π\Rightarrow x = 2\pi + \dfrac{\pi }{6} + \pi
Simplify the expression to find the second solution.
⇒x=2π+π6+π=19π6\Rightarrow x = 2\pi + \dfrac{\pi }{6} + \pi = \dfrac{{19\pi }}{6}
To find the second solution, subtract the solution from 2π2\pi , to find a reference angle
⇒x=19π6−2π=7π6\Rightarrow x = \dfrac{{19\pi }}{6} - 2\pi = \dfrac{{7\pi }}{6}
Add 2π2\pi to every negative angle to get positive angles.
⇒x=−π6+2π=11π6\Rightarrow x = - \dfrac{\pi }{6} + 2\pi = \dfrac{{11\pi }}{6}
The period of the sin(x)sin\left( x \right) function is 2π2\pi so values will repeat every 2π2\pi radians in both directions.
x=7π6+2πn,11π6+2πnx = \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=7π6,11π6x = \dfrac{{7\pi }}{6},\dfrac{{11\pi }}{6}
If n= 1, then value of x=19π6,23π6x = \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 4π4\pi .

So the four values of x for equation 2sin⁡x=−1,  0⩽x⩽4π2\sin x = - 1,\;0 \leqslant x \leqslant 4\pi are (7π6,11π6,19π6,23π6)\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 2π2\pi . In other words, every 2π2\pi units, the y-values repeat. If we need to find all possible solutions, then we must add 2πk2\pi k where k is an integer, to the initial solution.