Solve algebraically and show work please.
general solution and all solutions within 0,2pi
2sin(3x-pi/5) =1
To solve the equation 2sin(3x-pi/5) = 1 algebraically, we can start by isolating the sine term. Here's the step-by-step solution:
1. Start with the equation: 2sin(3x-pi/5) = 1
2. Divide both sides by 2: sin(3x-pi/5) = 1/2
3. Take the inverse sine (sin^(-1)) of both sides to eliminate the sine function: 3x-pi/5 = sin^(-1)(1/2)
4. Find the principal value of sin^(-1)(1/2): sin^(-1)(1/2) = pi/6
5. Solve for 3x-pi/5: 3x-pi/5 = pi/6
6. Add pi/5 to both sides: 3x = pi/6 + pi/5
7. Find a common denominator: 3x = 5pi/30 + 6pi/30
8. Combine the fractions: 3x = 11pi/30
9. Divide both sides by 3: x = (11pi/30) / 3
10. Simplify the expression: x = 11pi/90
Thus, the general solution for the equation 2sin(3x-pi/5) = 1 is x = 11pi/90.
To find all solutions within the interval [0, 2pi], we need to check for any additional solutions by adding multiples of 2pi to the general solution.
First, let's rewrite 11pi/90 in terms of the least common denominator (LCD) which is 90:
11pi/90 = (11pi/90) * (2pi/2pi) = (22pi^2)/(180pi)
Now, we can add multiples of 2pi to the general solution:
22pi^2/180pi + 2pi * n/180pi, where n is an integer.
Simplifying the expression gives:
22pi^2/180pi + 2npi/180pi = (22pi^2 + 2npi) / 180pi
Therefore, all solutions within the interval [0, 2pi] for the equation 2sin(3x-pi/5) = 1 are:
x = (22pi^2 + 2npi) / 180pi, where n is an integer.
To solve the equation 2sin(3x - π/5) = 1, we need to isolate the sin function and find the values of x that satisfy the equation. Let's solve it step-by-step.
Step 1: Divide both sides of the equation by 2:
sin(3x - π/5) = 1/2
Step 2: To eliminate the sine function, take the arcsine (inverse sine) of both sides of the equation:
arcsin(sin(3x - π/5)) = arcsin(1/2)
Note: Taking the arcsine of both sides allows us to cancel out the sine function, but keep in mind that it introduces additional solutions.
Step 3: Due to the periodicity of the sine function, the general solution will involve multiples of 2π. The arcsine, or sin^-1, has a range of -π/2 to π/2. Therefore, we need to consider multiple cases.
Case 1: arcsin(1/2) = π/6
When 3x - π/5 = π/6:
3x = π/6 + π/5
3x = (5π + 6π) / 30
3x = 11π / 30
x = (11π / 30) / 3
x = 11π / 90
Case 2: arcsin(1/2) = π - π/6 = 5π/6
When 3x - π/5 = 5π/6:
3x = 5π/6 + π/5
3x = (5π + 6π) / 6
3x = 11π / 6
x = (11π / 6) / 3
x = 11π / 18
Note: We have found two solutions so far, but remember to consider the periodicity of the sine function.
Step 4: Because the sine function is periodic with a period of 2π, we need to add or subtract multiples of 2π to our solutions to find all solutions within the interval [0, 2π].
To find additional solutions, let's add 2π to our previous solutions:
For x = 11π / 90 + 2π:
x ≈ 11π / 90 ≈ 0.387π
For x = 11π / 18 + 2π:
x ≈ 11π / 18 + 2π
x ≈ 21π / 18 ≈ 1.167π
Thus, the solutions within the interval [0, 2π] are approximately x ≈ 0.387π, x ≈ 1.167π.