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## Trigonometry

### Course: Trigonometry>Unit 4

Lesson 2: Sinusoidal equations

# Sine equation algebraic solution set

Solve a sine equation with an infinite number of solutions. Use trig identities to represent the whole solution set. Created by Sal Khan.

## Want to join the conversation?

• In the exercises, +2πn (or +360n) is added directly to the right side of the equation, but in this video, it is added first to x inside the brackets then later on transposed to the opposite side so it became -2πn (or -360n).

I got the same numbers but the only difference is the sign. For example, in one of the exercises, I got the solution: -6.07-45n following Sal's method. But in the choices, it only provides -6.07+45n. When I checked the explanation, the +360n is added directly on the right side. Can someone explain which way is the correct evaluation of the solution? • I dont get why on the exercises they say we use sin(180-O) but then in the process they use sin(-180 -O) • How and when do we differentiate between adding the 2pi(n) to the X side or the opposite side.

In previous lessons the 2pi(n) was added to the right side.

When I did the problem before watching this lesson I ended up with x= 4sin^-1(3/8+2pi(n)) and x=-4sin^-1(3/8-2pi(n)) as I thought you just add that bit in there but now I am getting confused as when to know exactly how and when to add this step.. (the 2pi(n) and such) • When solving trigonometric equations, it is important to remember that the trigonometric functions are periodic, meaning they repeat themselves after a certain interval. For example, the sine function has a period of 2π, which means that sin(x) = sin(x + 2π) for any value of x.

When solving trigonometric equations, we may need to find all solutions that satisfy the equation. To do this, we often use the general solution, which includes all possible solutions, and then use the periodicity of the trigonometric functions to find all solutions in a given interval.

When using the general solution, we add 2πn (or a multiple of the period) to the angle on the right-hand side of the equation. This is because adding 2πn to the angle will give us another angle that has the same trigonometric function value.

For example, if we have the equation sin(x) = 0.5, we can use the inverse sine function to find one solution: x = sin^-1(0.5) = π/6. But since the sine function has a period of 2π, we know that there are other angles that have the same sine value, such as x = 5π/6, 13π/6, etc. To find all solutions, we use the general solution:

``x = π/6 + 2πn or x = 5π/6 + 2πn``

where n is an integer. This gives us all possible solutions to the equation.

So, to answer your question, we add the 2πn to the angle on the right-hand side of the equation when using the general solution to find all solutions to a trigonometric equation. It is important to keep track of which quadrant each solution is in, as this will affect the sign of the trigonometric function.
• At , Sal takes the inverse sine of both sides of the equation. He assumes that arcsin(sin(...))=... on the left side.
However, as mentioned in the previous videos, the range of the arcsine function is -π/2 to π/2. This means that if you take arcsin(sin(θ)), you don't always get the result θ.
I understand Sal intends to get the complete solution set, but I wonder if he is being mathematically rigorous at this. • In the exercises, it asked me to find all solutions to a sine equation like the one in this video and I did get the sin(θ+2πn) answer correct but I couldn't find the sin(π-θ+2πn) answer in the provided answers, and under the hints it says that we only find the answers in the -π ≤ θ ≤ π interval, and that the first answer was the "trough" and the only correct one. I would appreciate any help that explains this. • If the first solution was either 𝜃 = 𝜋∕2 or 𝜃 = −𝜋∕2, then there is no integer 𝑛 for which 𝜋 − 𝜃 + 2𝜋𝑛 is a different solution in the interval [−𝜋, 𝜋].

In mathematics a trough is basically the bottom of a curve, in this case the bottom of the unit circle, i.e. 𝜃 = −𝜋∕2   