solve 2sin^2x+sinx-1=0 in the interval [0, 2pi0
Call sin a new variable y and solve the quadratic equation for y:
2y^2 +y -1 = 0
(2y - 1) (y + 1) = 0
y = sin x = -1/2 or -1
To solve the equation 2sin^2x + sinx - 1 = 0 in the interval [0, 2π], we can use the quadratic formula. But first, let's simplify the equation a bit.
Given equation: 2sin^2x + sinx - 1 = 0
Let's substitute sin^2x as (1 - cos^2x):
2(1 - cos^2x) + sinx - 1 = 0
Expanding and rearranging the terms:
2 - 2cos^2x + sinx - 1 = 0
-2cos^2x + sinx + 1 = 0
Now we can use the quadratic formula to solve for cosx:
cosx = (-b ± √(b^2 - 4ac)) / (2a)
a = -2, b = 1, c = 1
cosx = (-(1) ± √((1)^2 - 4(-2)(1))) / (2(-2))
cosx = (-1 ± √(1 + 8)) / (-4)
cosx = (-1 ± √9) / (-4)
cosx = (-1 ± 3) / (-4)
There are two possible solutions for cosx:
1. When cosx = (-1 + 3) / (-4) = 2 / -4 = -0.5
2. When cosx = (-1 - 3) / (-4) = -4 / -4 = 1
To find the corresponding values of x, we can use the inverse cosine function:
1. When cosx = -0.5, x = π ± arccos(-0.5)
2. When cosx = 1, x = π ± arccos(1)
For the interval [0, 2π], the values of x will be:
1. When cosx = -0.5:
x = π ± arccos(-0.5) ≈ π ± 2.094 (approximately)
Since these two values are outside the interval [0, 2π], we discard them.
2. When cosx = 1:
x = π ± arccos(1) ≈ π ± 0
So, the solutions for the equation 2sin^2x + sinx - 1 = 0 in the interval [0, 2π] are:
x = π, when cosx = 1
Therefore, the solution to the equation is x = π.
To solve the equation 2sin^2x + sinx - 1 = 0 within the interval [0, 2pi], we can use the quadratic formula.
First, let's rewrite the equation in terms of a quadratic equation form. Notice that sin^2x is equivalent to (sinx)^2.
So, our equation becomes:
2(sin^2x) + sinx - 1 = 0
Let's substitute y = sinx:
2y^2 + y - 1 = 0
Now we can use the quadratic formula, which states that for a quadratic equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
In our case, a = 2, b = 1, and c = -1. Plugging in these values, we have:
y = (-1 ± sqrt(1^2 - 4(2)(-1))) / (2(2))
y = (-1 ± sqrt(1 + 8)) / 4
y = (-1 ± sqrt(9)) / 4
y = (-1 ± 3) / 4
We have two possible values for y:
y = (2) / 4 = 1/2
y = (-4) / 4 = -1
Now that we have the values of y, we can substitute back sinx for y:
sinx = 1/2 or sinx = -1
To find the solutions within the interval [0, 2pi], we need to determine the values of x that correspond to these solutions.
For sinx = 1/2, the values of x in the interval [0, 2pi] are π/6 and 5π/6. These correspond to the angles where sinx equals 1/2.
For sinx = -1, the value of x in the interval [0, 2pi] is 3π/2. This corresponds to the angle where sinx equals -1.
Therefore, the solutions to the equation 2sin^2x + sinx - 1 = 0 in the interval [0, 2pi] are:
x = π/6, 5π/6, and 3π/2.