Find general solution of
sin(pi-6x)+root 3 sin((pi/2) +6x))=root3
sin(π-6x) = sin6x
sin(π/2 + 6x) = cos6x
so now you have
sin6x + √3 cos6x = √3
√3/2 cos6x + 1/2 sin6x = √3/2
cos(6x - π/2) = √3/2
6x - π/2 = π/6 or -π/6
6x = (2π/3 or π/3) + 2kπ
x = (π/9 or π/18) +kπ/3
To find the general solution of the equation sin(pi - 6x) + √3sin((π/2) + 6x) = √3, we'll solve it step by step.
First, let's simplify the equation by using trigonometric identities:
1. sin(pi - 6x) = sin(pi)cos(6x) - cos(pi)sin(6x) = 0 -(-1)sin(6x) = sin(6x)
2. √3sin((π/2) + 6x) = √3cos(-6x) = √3cos(6x)
After simplifying the equation, it becomes:
sin(6x) + √3cos(6x) = √3
Next, we can rewrite this equation in terms of a single trigonometric function using the sine and cosine addition formulas:
Let's define: A = sin(6x), B = cos(6x)
sin(6x) + √3cos(6x) = √3
A + √3B = √3
Now we can square both sides of the equation:
(A + √3B)^2 = (√3)^2
A^2 + 2√3AB + 3B^2 = 3
Expanding and simplifying:
A^2 + 2√3AB + 3B^2 - 3 = 0
Now, we can use the quadratic formula to solve for A and B:
A = [-2√3B ± √(4√3^2B^2 - 4(3)(B^2 - 3))]/(2)
A = [-2√3B ± √(12B^2 - 12B^2 + 36)]/2
A = [-2√3B ± √36]/2
A = [-2√3B ± 6]/2
A = -√3B ± 3
Now, we can substitute A and B back into the equation:
sin(6x) = -√3cos(6x) ± 3
We can rewrite this as a single trigonometric function:
tan(6x) = -√3 ± 3/√3
Now we can solve for 6x:
6x = atan(-√3 ± 3/√3)
Finally, to find the general solution, we can add multiples of 2π to each term:
x = (atan(-√3 ± 3/√3) + 2πn)/6
where n is an integer representing all possible solutions.
To find the general solution of the equation sin(pi - 6x) + √3 sin((π/2) + 6x) = √3, we can start by simplifying the equation.
Using the trigonometric identity sin(π - θ) = sin θ and sin(π/2 + θ) = cos θ, we can rewrite the equation as follows:
sin(6x) + √3 cos(6x) = √3
Next, we can square both sides of the equation to eliminate the square root:
[ sin(6x) + √3 cos(6x) ]^2 = (√3)^2
sin^2(6x) + 2sin(6x)√3cos(6x) + 3cos^2(6x) = 3
Using the trigonometric identity sin^2(θ) + cos^2(θ) = 1, we can simplify further:
1 - cos^2(6x) + 2sin(6x)√3cos(6x) + 3cos^2(6x) = 3
Rearranging the terms, we have:
4cos^2(6x) - 2sin(6x)√3cos(6x) - 2 = 0
Now, let's solve this quadratic equation for cos(6x):
Using the quadratic formula:
cos(6x) = [ -(-2sin(6x)√3) ± √((-2sin(6x)√3)^2 - 4(4)(-2))] / (2(4))
cos(6x) = [2sin(6x)√3 ± √(12sin^2(6x) - 32)] / 8
cos(6x) = [sin(6x)√3 ± √(3sin^2(6x) - 8)] / 4
To find the values of sin(6x) that satisfy this equation, we can use the fact that sin^2(θ) + cos^2(θ) = 1.
Since cos(6x) = √[1 - sin^2(6x)], we can substitute this into the equation:
√[1 - sin^2(6x)] = [sin(6x)√3 ± √(3sin^2(6x) - 8)] / 4
Squaring both sides of the equation to eliminate the square root gives:
1 - sin^2(6x) = [sin^2(6x)*3 ± 2sin(6x)√(3sin^2(6x) - 8) + 3sin^2(6x) - 8] / 16
Multiplying through by 16 gives:
16 - 16sin^2(6x) = 3sin^2(6x) ± 2sin(6x)√(3sin^2(6x) - 8) + 3sin^2(6x) - 8
Combining like terms gives:
19sin^2(6x) ± 2sin(6x)√(3sin^2(6x) - 8) - 8 = 16
Rearranging the terms gives:
19sin^2(6x) ± 2sin(6x)√(3sin^2(6x) - 8) = 24
Now, we have a quadratic equation in terms of sin(6x). We can solve this equation using factoring or the quadratic formula to find the values of sin(6x). Once we have the values of sin(6x), we can find the values of cos(6x) using the equation cos(6x) = √[1 - sin^2(6x)].
Finally, the general solution for x will be the values of x that satisfy sin(6x) and cos(6x) for the given values of sin(6x) and cos(6x).