Asksia AI LOGO

Sia

Question
Math
Posted 4 months ago

Select one or more expressions that together represent all solutions to the equation. Your answer should be in radians.
Assume n\boldsymbol{n} is any integer.
13sin⁡(2x)−3=3
13 \sin (2 x)-3=3


Choose all answers that apply:

A −0.480+n⋅2π-0.480+n \cdot 2 \pi

B −0.240+n⋅π-0.240+n \cdot \pi
c 0.240+n⋅π0.240+n \cdot \pi
D 1.331+n⋅π1.331+n \cdot \pi

E 0.480+n⋅2π0.480+n \cdot 2 \pi

F 2.661+n⋅2π2.661+n \cdot 2 \pi
Sign in to unlock the answer
Answer from Sia
Posted 4 months ago
Solution by Steps
step 2
Adding 3 to both sides of the equation gives us 13sin⁡(2x)=613 \sin(2x) = 6
step 3
Dividing both sides by 13 gives us sin⁡(2x)=613\sin(2x) = \frac{6}{13}
step 4
To find xx, we take the inverse sine of 613\frac{6}{13}, which gives us 2x=sin⁡−1(613)2x = \sin^{-1}\left(\frac{6}{13}\right)
step 5
The general solution for xx in terms of nn is x=12(2πn+π−sin⁡−1(613))x = \frac{1}{2} \left(2 \pi n + \pi - \sin^{-1}\left(\frac{6}{13}\right)\right) and x=πn+12sin⁡−1(613)x = \pi n + \frac{1}{2} \sin^{-1}\left(\frac{6}{13}\right) for n∈Zn \in \mathbb{Z}
step 6
We calculate the value of 12sin⁡−1(613)\frac{1}{2} \sin^{-1}\left(\frac{6}{13}\right) to find the specific solutions that match the given options
step 7
The value of sin⁡−1(613)\sin^{-1}\left(\frac{6}{13}\right) is approximately 0.4800.480 radians. Therefore, the solutions are x=πn+0.240x = \pi n + 0.240 and x=πn−0.240x = \pi n - 0.240
step 8
Comparing these solutions to the given options, we find that option C and option B represent all solutions to the equation
B, C
Key Concept
General Solution for Trigonometric Equations
Explanation
The general solution for a trigonometric equation of the form sin⁡(θ)=a\sin(\theta) = a is θ=sin⁡−1(a)+2πn\theta = \sin^{-1}(a) + 2\pi n or θ=π−sin⁡−1(a)+2πn\theta = \pi - \sin^{-1}(a) + 2\pi n, where nn is any integer. For the equation sin⁡(2x)=613\sin(2x) = \frac{6}{13}, we find the specific solutions and express them in a form that matches the given multiple-choice options.

Not the question you are looking for? Ask here!

Enter question by text

Enter question by image

Upgrade to Asksia Pro

Join a AskSia's Pro Plan, and get 24/7 AI tutoring for your reviews, assignments, quizzes and exam preps.

Unlimited chat query usages
Strong algorithms that better know you
Early access to new release features
Study Other Question