Find all solutions of the equation 2sin^2x-cosx=1 in the interval [0,2pi)
x1= ?
x2=?
x3=?
2 (1 - cos^2 x) - cos x = 1
2 - 2 cos^2 x -cos x = 1
2 cos^2 x + cos x -1 = 0
let y = cos x
then
2 y^2 +y -1 = 0
(y+1)(2y-1) = 0
y = .5 or y = -1
so
cos x = .5 or cos x = -1
so
x = 60 degrees or 300 degrees
or 180 degrees
Rewrite as
2(1-cos^2x)-cosx = 1.
Rearrange as
2cos^2x + cos x -1 = 0,
which can be factored to provide:
(2cosx -1)(cosx +1) = 0.
This means
cosx = 1/2 or -1
Use that to find the three angles.
To find all solutions of the equation 2sin^2x - cosx = 1 in the interval [0, 2π), we can solve it step-by-step:
Step 1: Let's rearrange the equation to make it easier to solve:
2sin^2x - cosx - 1 = 0
Step 2: Notice that 2sin^2x - cosx - 1 can be factored as follows:
(2sinx + 1)(sinx - 1) = 0
Step 3: Now, set each factor equal to zero and solve separately:
2sinx + 1 = 0
sinx - 1 = 0
For the first factor:
2sinx + 1 = 0
2sinx = -1
sinx = -1/2
To find the solutions for sinx = -1/2 in the interval [0, 2π), we can use the unit circle or trigonometric values. The solutions are:
x1 = 7π/6
x2 = 11π/6
For the second factor:
sinx - 1 = 0
sinx = 1
To find the solutions for sinx = 1 in the interval [0, 2π), we again use the unit circle or trigonometric values. The solutions are:
x3 = π/2
Therefore, the solutions to the equation 2sin^2x - cosx = 1 in the interval [0, 2π) are:
x1 = 7π/6
x2 = 11π/6
x3 = π/2
To find all the solutions of the equation 2sin^2x - cosx = 1 in the interval [0, 2π), we can follow these steps:
Step 1: Rewrite the equation in terms of only sines or cosines.
Let's use the identity sin^2x = 1 - cos^2x to express sin^2x in terms of cosine:
2(1 - cos^2x) - cosx = 1
Expanding and rearranging, we get: 2cos^2x + cosx - 1 = 0
Step 2: Solve the quadratic equation.
To solve this quadratic equation, we can factor it or use the quadratic formula. In this case, factoring is easier:
(2cosx - 1)(cosx + 1) = 0
Setting each factor equal to zero:
Factor 1: 2cosx - 1 = 0
Solving for cosx gives: cosx = 1/2
This leads to two solutions within the interval: x = π/3 or x = 5π/3
Factor 2: cosx + 1 = 0
Solving for cosx gives: cosx = -1
This leads to one solution within the interval: x = π
Step 3: Finalize the solutions within the given interval.
The solutions we found (π/3, 5π/3, and π) all fall within the interval [0, 2π). Therefore:
x1 = π/3
x2 = 5π/3
x3 = π
So, the solutions to the equation 2sin^2x - cosx = 1 in the interval [0, 2π) are x1 = π/3, x2 = 5π/3, and x3 = π.