We will solve the given equation by first isolating

sin θ

on the left-hand side.

2 sin θ + 1 = 0

▼

Solve for

sin θ

SubEqn

$LHS - 1 = RHS - 1$

2 sin θ = - 1

DivEqn

$LHS / 2 = RHS / 2$

sin θ = \frac{- 1}{2}

MoveNegNumToFrac

Put minus sign in front of fraction

sin θ = - \frac{1}{2}

Next, we will apply the inverse of the sine function on both sides of the equation.

sin θ = - \frac{1}{2}

$sin^{- 1} (LHS) = sin^{- 1} (RHS)$

θ = sin^{- 1} - \frac{1}{2}

UseCalc

Use a calculator

θ = - 3 0^{\circ}

We found that one solution to the equation is

θ = - 3 0^{\circ} .

However, we want the solutions to be in the interval from

0^{\circ}

to

36 0^{\circ} .

Since the solution we found is negative, the angle is measured clockwise. To find the solution in the desired interval, we subtract

30

from

360 .

The first solution in the desired interval is $θ = 33 0^{\circ} .$ To find other possible solutions, recall that the sine of an angle is the second coordinate of the point of intersection between the unit circle and the terminal side of the angle. Therefore, we will find another angle whose terminal side intersects the unit circle at a point with second coordinate $- \frac{1}{2} .$

In the diagram, we can see that another solution to the equation is

θ = 21 0^{\circ} .

In fact, we see that, in the desired interval, the only solutions to the equation are

θ = 21 0^{\circ}

and

θ = 33 0^{\circ} .

Checking Our Answer

We can check our answer by graphing $y = 2 sin θ + 1$ and $y = 0$ in the same coordinate plane on a graphic calculator. To do so, we must set the calculator in degree mode. We do this by pushing $M O D E$ and selecting Degree instead of Radian in the third row.

We want to find solutions between $0^{\circ}$ and $36 0^{\circ} .$ Accordingly, let's resize the window of the calculator to show $x -$ values between $0$ and $360,$ and $y -$ values between $- 4$ and $4 .$ and To do so, we push $WINDOW$ and change the settings.

Now, let's graph both functions. The $x -$ coordinate of the points of intersection, if any, will be the solutions to the equation. Press the $Y =$ button and type the functions in the first two rows. Having written the functions, push $GRAPH$ to draw them.

We can see that there are two points of intersection. To find them, push $2 nd$ and $CALC$ and choose the fifth option, which is intersect.

Next, we need to choose left and right boundaries for one of the points. Finally, the calculator asks for a guess where the intersection point might be. After that, it will calculate the exact point for us. We have to do this twice, once for each point.

We see that the two solutions match the ones we have previously found.

Exercise

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