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.
Solve algebraically and show work please.
general solution and all solutions within 0,2pi
2sin(3x-pi/5) =1
1 answer