Question 2 of 20
This test: 20 point(s) possible
This question: 1 point(s) possible

Find one solution for the equation. Assume that all angles involved are acute angles.

(Simplify your answer.)

Ask by Long Hamilton. in the United States

Mar 13,2025

Question

Question 2 of 20
This test: 20 point(s) possible
This question: 1 point(s) possible

Find one solution for the equation. Assume that all angles involved are acute angles.

(Simplify your answer.)

Ask by Long Hamilton. in the United States

Mar 13,2025

UpStudy AI · Accepted Answer

To solve the equation

$$
\sec \left(3 \beta + 16^{\circ}\right) = \csc \left(\beta - 2^{\circ}\right),
$$

we start by recalling the definitions of secant and cosecant:

$$
\sec(x) = \frac{1}{\cos(x)} \quad \text{and} \quad \csc(y) = \frac{1}{\sin(y)}.
$$

Thus, we can rewrite the equation as:

$$
\frac{1}{\cos(3\beta + 16^{\circ})} = \frac{1}{\sin(\beta - 2^{\circ})}.
$$

Cross-multiplying gives us:

$$
\sin(\beta - 2^{\circ}) = \cos(3\beta + 16^{\circ}).
$$

Using the co-function identity $$\sin(x) = \cos(90^{\circ} - x)$$, we can rewrite the equation as:

$$
\cos(90^{\circ} - (\beta - 2^{\circ})) = \cos(3\beta + 16^{\circ}).
$$

This simplifies to:

$$
\cos(92^{\circ} - \beta) = \cos(3\beta + 16^{\circ}).
$$

Since the cosine function is periodic, we have two cases to consider:

1. $$92^{\circ} - \beta = 3\beta + 16^{\circ} + k \cdot 360^{\circ}$$
2. $$92^{\circ} - \beta = - (3\beta + 16^{\circ}) + k \cdot 360^{\circ}$$

where $$k$$ is any integer.

Let's solve the first case:

$$
92^{\circ} - \beta = 3\beta + 16^{\circ}.
$$

Rearranging gives:

$$
92^{\circ} - 16^{\circ} = 4\beta,
$$
$$
76^{\circ} = 4\beta,
$$
$$
\beta = \frac{76^{\circ}}{4} = 19^{\circ}.
$$

Now, let's check the second case:

$$
92^{\circ} - \beta = - (3\beta + 16^{\circ}).
$$

Rearranging gives:

$$
92^{\circ} - \beta = -3\beta - 16^{\circ},
$$
$$
92^{\circ} + 16^{\circ} = -2\beta,
$$
$$
108^{\circ} = -2\beta,
$$
$$
\beta = -54^{\circ}.
$$

Since we are looking for acute angles, we discard $$-54^{\circ}$$.

Thus, the only solution for $$\beta$$ is:

$$
\beta = 19^{\circ}.
$$

So, the final answer is:

$$
\beta = 19^{\circ}.
$$
$$ \beta = 19^{\circ} $$

Question 2 of 20
This test: 20 point(s) possible
This question: 1 point(s) possible

Find one solution for the equation. Assume that all angles involved are acute angles.

(Simplify your answer.)

Solución de inteligencia artificial de Upstudy

Responder

Solución

Mind Expander

Solving the first equation:

Solving the second equation:

preguntas relacionadas

Latest Trigonometry Questions

Question 2 of 20 This test: 20 point(s) possible This question: 1 point(s) possible Find one solution for the equation. Assume that all angles involved are acute angles. (Simplify your answer.)

Solución de inteligencia artificial de Upstudy

Responder

Solución

Mind Expander

Solving the first equation:

Solving the second equation:

preguntas relacionadas

Latest Trigonometry Questions

Question 2 of 20
This test: 20 point(s) possible
This question: 1 point(s) possible

Find one solution for the equation. Assume that all angles involved are acute angles.

(Simplify your answer.)