solve Sinx=Cos2x-1 for 0¡Üx<2¦Ð
To solve the equation sin(x) = cos(2x) - 1 for the given range 0 ≤ x < 2π, we can use a combination of algebraic manipulation and trigonometric identities.
Let's start by simplifying the equation.
sin(x) = cos(2x) - 1
Next, we can use the double-angle identity for cosine, which states that cos(2x) = 1 - 2sin^2(x). Substituting this into the equation, we get:
sin(x) = 1 - 2sin^2(x) - 1
Simplifying further, we have:
sin(x) = - 2sin^2(x)
Now, let's move all the terms to one side of the equation:
2sin^2(x) + sin(x) = 0
Factoring out sin(x), we get:
sin(x)(2sin(x) + 1) = 0
Now, let's solve each factor separately.
1) sin(x) = 0:
This gives us the solutions x = 0 and x = π.
2) 2sin(x) + 1 = 0:
Subtracting 1 from both sides, we have:
2sin(x) = -1
Dividing both sides by 2, we get:
sin(x) = -1/2
This equation has solutions in two quadrants: the third and fourth quadrants. In the third quadrant, sin(x) = -1/2, so x = (7π)/6. In the fourth quadrant, sin(x) = -1/2, so x = (11π)/6.
Therefore, the solutions for the equation sin(x) = cos(2x) - 1 in the range 0 ≤ x < 2π are:
x = 0, x = π, x = (7π)/6, and x = (11π)/6.