We can find the solutions of

sin x = 0.6
algebraically. (Round your answers to two decimal places.)

(a) First we find the solutions in the interval
[0, 2π).
We get one such solution by taking
sin−1
to get
x = ________
(smaller value).

The other solution in this interval is
x =
(larger value).


(b) We find all solutions by adding multiples of
to the solutions in
[0, 2π).
The solutions are
x =
.
(Enter your answers in the form θ + 2πk, 0 ≤ θ < 2π. Enter your answers as a comma-separated list.)

0.64, 2.50 (π-0.64)

I'm not sure what is meant by part (b), but in the given domain, the only two answers are 0.64 and 2.50 (2 decimal places)

(a) First, let's find the solution in the interval [0, 2π) by taking sin^-1 of 0.6:

sin^-1(0.6) ≈ 0.64

Therefore, the smaller value in the interval [0, 2π) is x = 0.64.

To find the larger value, we can use the fact that sine has a periodic nature. Since sin(0) = sin(2π) = 0 and sin is an odd function, we can observe that sin(x) = 0.6 has symmetrical solutions around the point x = π.

So, the larger value in the interval [0, 2π) is:

x = π - 0.64 ≈ 2.50

(b) To find all the solutions by adding multiples of π to the solutions in [0, 2π), we have:

x = 0.64 + 2πk, 0 ≤ θ < 2π, where k is an integer.

Therefore, the solutions are:

x = 0.64 + 2πk, where k is an integer.

So, the solutions are x = 0.64, 0.64 + 2π, 0.64 + 4π, 0.64 + 6π, ... (entering as a comma-separated list).

Note: Clown Bot would be happy to entertain you while answering your questions, but please remember that the answers provided are for educational purposes only.

To find the solutions of sin x = 0.6 algebraically, we can follow these steps:

(a) First we find the solutions in the interval [0, 2π).

Step 1: Take the inverse sine (sin^(-1)) of both sides of the equation:

sin^(-1)(sin x) = sin^(-1)(0.6)

This simplifies to:

x = sin^(-1)(0.6)

Step 2: Use a calculator to find the inverse sine of 0.6. Round the answer to two decimal places.

x = 0.93 (smaller value)

Step 3: To find the larger value, we can use the fact that sine function is periodic with a period of 2π. Since x = 0.93 is in the interval [0, 2π), we can add 2π to get the next solution:

x = 0.93 + 2π

Now we can round this to two decimal places:

x ≈ 6.05 (larger value)

(b) To find all solutions, we add multiples of π to the solutions in [0, 2π).

The solutions are:

x = 0.93 + 2πk, 0 ≤ x < 2π

So the final answer is x ≈ 0.93 + 2πk, where k is an integer, and the solutions are presented as a comma-separated list.

To find the solutions of the equation sin(x) = 0.6 algebraically, we can follow these steps:

(a) In the interval [0, 2π), we can find the solutions as follows:

Step 1: Take the inverse sine function (sin^(-1)) of both sides of the equation to isolate x:
sin^(-1)(sin(x)) = sin^(-1)(0.6)

Step 2: Simplify the expression on the left-hand side using the inverse sine identity:
x = sin^(-1)(0.6)

Step 3: Evaluate sin^(-1)(0.6) using a calculator or reference table. The answer will be in radians. Round the value to two decimal places.
x = 0.6435 (smaller value)

Step 4: Since sine is a periodic function with a period of 2π, we need to find the other solution in the interval [0, 2π). By symmetry, the other solution will be π - x:
x = π - 0.6435
x = 2.4981 (larger value)

Therefore, the solutions in the interval [0, 2π) are x = 0.6435 (smaller value) and x = 2.4981 (larger value).

(b) To find all solutions by adding multiples of π to the solutions in [0, 2π), we can use the following formula:

x = θ + 2πk

where θ represents the solutions found in part (a), and k is an integer.

Using this formula, we can express the solutions as:
x = 0.6435 + 2πk, 0 ≤ θ < 2π
x = 2.4981 + 2πk, 0 ≤ θ < 2π

These are the general solutions to the equation sin(x) = 0.6, expressed in the form θ + 2πk, where k can be any integer.