Solve Sinx=Cos2x-1 for all values between 0 and 2pi
cos 2x = 1 - 2sin^2 x
so, you have
2sin^2 x + sin x = 0
sin x (2sin x + 1) = 0
sinx = 0 --> x = 0,2pi
sinx = -1/2 --> x = 7pi/6,11pi/6
To solve the equation sin(x) = cos(2x) - 1, we need to find the values of x that satisfy this equation within the interval [0, 2π].
To solve this equation, we'll start by bringing all the terms to one side:
sin(x) - (cos(2x) - 1) = 0
Next, we'll use some trigonometric identities to simplify the equation. The double-angle identity for cosine states that:
cos(2x) = 2cos^2(x) - 1
Applying this identity to our equation, we get:
sin(x) - (2cos^2(x) - 1) - 1 = 0
Simplifying further, we have:
sin(x) - 2cos^2(x) + 2 = 0
Now, let's use the Pythagorean identity: sin^2(x) + cos^2(x) = 1
Rearranging this identity, we get: sin^2(x) = 1 - cos^2(x)
Substituting sin^2(x) in terms of cos^2(x) in our equation, we have:
1 - cos^2(x) - 2cos^2(x) + 2 = 0
Combining like terms, we get:
-3cos^2(x) - cos^2(x) + 3 = 0
Now, let's simplify further:
-4cos^2(x) + 3 = 0
Adding 4cos^2(x) to both sides, we have:
4cos^2(x) = 3
Finally, dividing both sides by 4, we get:
cos^2(x) = 3/4
Taking the square root of both sides, we have:
cos(x) = ±√(3/4)
cos(x) = ±√(3)/2
Now, to find the values of x, we need to consider two cases:
1) cos(x) = √(3)/2
2) cos(x) = -√(3)/2
Case 1: cos(x) = √(3)/2
To find the values of x for which cos(x) = √(3)/2, we can refer to the unit circle or the cosine function's values on the unit circle. The cosine function is positive in the first and fourth quadrants of the unit circle.
In the first quadrant, x = π/6.
In the fourth quadrant, x = (11π/6).
Case 2: cos(x) = -√(3)/2
To find the values of x for which cos(x) = -√(3)/2, we can again refer to the unit circle or the cosine function's values on the unit circle. The cosine function is negative in the second and third quadrants of the unit circle.
In the second quadrant, x = (5π/6).
In the third quadrant, x = (7π/6).
Therefore, the solutions to the equation sin(x) = cos(2x) - 1 within the interval [0, 2π] are:
x = π/6, (5π/6), (7π/6), (11π/6).