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Solve the following for
0 < Theta < 360


sin theta + 3/4 = 0

User Bobsickle
by
7.0k points

2 Answers

3 votes

Answer:

θ = 311.41°

θ = 228.59°

Explanation:

To solve the trigonometric equation sin(θ) + 3/4 = 0 for the given interval 0° < θ < 360°, we can use algebraic manipulation and inverse trigonometric functions.

Isolate sin(θ) by subtracting 3/4 from both sides of the equation:


\begin{aligned}\sin \theta+(3)/(4)&amp;=0\\\\\sin \theta+(3)/(4)-(3)/(4)&amp;=0-(3)/(4)\\\\\sin \theta &amp;= -(3)/(4)\end{aligned}

Take the inverse sine (arcsin) of both sides to solve for θ:


\begin{aligned}\arcsin \left(\sin \theta\right) &amp;=\arcsin \left( -(3)/(4)\right)\\\\\theta&amp;=-48.5903778...^(\circ)\end{aligned}

A positive angle is measured counterclockwise from the positive x-axis, whereas a negative angle is measured clockwise from the positive x-axis. Therefore, θ is in quadrant IV.

As sine is negative in quadrants III and IV, the other value of θ when sinθ = -3/4 is in quadrant III. Therefore, to find the other solution, add 180° to the initial solution:


\theta=-48.5930778...^(\circ) + 180^(\circ) = 228.5903778...^(\circ)

The sine function is periodic with a period of 360°. Therefore, the solutions are:


\theta = -48.5930778...^(\circ) + 360^(\circ)n


\theta = 228.5903778...^(\circ) + 360^(\circ)n

Therefore, the solutions for the given equation in the interval 0° < θ < 360° are:


\theta = -48.5930778...^(\circ) + 360^(\circ) = 311.41^(\circ) \; \sf(2\;d.p.)


\theta = 228.5903778...^(\circ) = 228.59^(\circ) \; \sf(2\;d.p.)

User Smilingpoplar
by
7.6k points
0 votes

Answer:


\tt \Theta \approx 228.59^\circ \quad \text{and} \quad \Theta \approx 311.41^\circ

Explanation:

Let's solve the trigonometric equation for
\tt 0 < \theta < 360^\circ


\tt \sin(\Theta) + (3)/(4) = 0

We can follow these steps:

Subtract
(3)/(4) from both sides to isolate sin θ.


\tt \sin(\Theta) = -(3)/(4)

Find the inverse sine (or arcsin) of both sides to find the possible values of θ.


\tt \Theta = \arcsin\left(-(3)/(4)\right)

Using a calculator to compute the arcsin and find the principal value of θ.

(since we're looking for values in the range
\tt 0 < \Theta < 360^\circ)


\tt \Theta \approx -48.59^\circ \quad

Since
\tt \sin(\Theta) is negative in both the third and forth quadrants, we can find the other solutions by adding 180° to the principal value:


\tt \Theta = 180^\circ + 48.59^\circ \approx 228.59^\circ


\tt \Theta = 360^\circ - 48.59 ^\circ \approx 311.41^\circ

Therefore,

The solutions for
0 < \Theta < 360^\circ are approximately:


\tt \Theta \approx 228.59^\circ \quad \text{and} \quad \Theta \approx 311.41^\circ

User Darrin
by
8.1k points

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