Question 7.6.1: Solving a Trigonometric Equation (Linear Methods) Solve the ...
Solving a Trigonometric Equation (Linear Methods)
Solve the equation 2 \sin θ + 1 = 0
(a) over the interval [0°, 360°) (b) for all solutions.
ALGEBRAIC SOLUTION
(a) Because \sin θ is to the first power, we use the same method as we would to solve the linear equation
2x + 1 = 0.
2 \sin θ + 1 = 0 Original equation
2 \sin θ = -1 Subtract 1.
\sin θ = – \frac{1}{ 2} Divide by 2.
To find values of θ that satisfy \sin θ = – \frac{1}{2} , we observe that θ must be in either quadrant III or quadrant IV because the sine function is negative only in these two quadrants. Furthermore, the reference angle must be 30°. The graph of the unit circle in Figure 29 shows the two possible values of θ. The solution set is {210°, 330°}.
(b) To find all solutions, we add integer multiples of the period of the sine function, 360°, to each solution found in part (a). The solution set is written as follows.
{210° + 360°n, 330° + 360°n,
where n is any integer}
GRAPHING CALCULATOR SOLUTION
(a) Consider the original equation.
2 \sin θ + 1 = 0
We can find the solution set of this equation by graphing the function
y_{1} = 2 \sin x + 1
and then determining its zeros. Because we are finding solutions over the interval [0°, 360°), we use degree mode and choose this interval of values for the input x on the graph.
The screen in Figure 30(a) indicates that one solution is 210°, and the screen in Figure 30(b) indicates that the other solution is 330°. The solution set is {210°, 330°}, which agrees with the algebraic solution.
(b) Because the graph of
y_{1} = 2 \sin x + 1
repeats the same y-values every 360°, all solutions are found by adding integer multiples of 360° to the solutions found in part (a). See the algebraic solution.
