2sin^2x+3cosx=3 find all solutions in the interval [0,2pi]
To find all solutions to the equation 2sin^2(x) + 3cos(x) = 3 in the interval [0, 2pi], we'll first rearrange the equation to a more convenient form:
2sin^2(x) + 3cos(x) - 3 = 0
Next, we'll use the trigonometric identity sin^2(x) + cos^2(x) = 1 to rewrite the equation:
2(1 - cos^2(x)) + 3cos(x) - 3 = 0
Simplifying further, we have:
2 - 2cos^2(x) + 3cos(x) - 3 = 0
Simplifying one more time:
-2cos^2(x) + 3cos(x) - 1 = 0
Now, let's solve this quadratic equation for cos(x) by factoring or using the quadratic formula. After obtaining the values for cos(x), we can substitute them back into the original equation to solve for x.
Using the quadratic formula:
cos(x) = [-b ± sqrt(b^2 - 4ac)] / 2a
For our equation, a = -2, b = 3, and c = -1. Plugging these values into the quadratic formula:
cos(x) = [-(3) ± sqrt((3)^2 - 4(-2)(-1))] / 2(-2)
cos(x) = [-3 ± sqrt(9 - 8)] / -4
cos(x) = [-3 ± sqrt(1)] / -4
cos(x) = (-3 ± 1) / -4
We have two possible solutions:
1. cos(x) = (-3 + 1) / -4 = -1 / -4 = 1/4
2. cos(x) = (-3 - 1) / -4 = -4 / -4 = 1
Now, let's substitute each value of cos(x) back into the original equation and solve for x.
1. cos(x) = 1/4:
2sin^2(x) + 3(1/4) = 3
2sin^2(x) + 3/4 = 3
2sin^2(x) = 3 - 3/4
2sin^2(x) = 9/4 - 3/4
2sin^2(x) = 6/4
sin^2(x) = 3/4
Taking the square root of both sides:
sin(x) = ±√3 / 2
The possible values for sin(x) are:
a) sin(x) = √3 / 2
b) sin(x) = -√3 / 2
To determine the values of x in the interval [0, 2pi] for each value of sin(x), we need to consider the quadrants in which sin(x) is positive or negative.
a) sin(x) = √3 / 2:
In the first and second quadrants, sin(x) is positive.
Taking the inverse sine:
x = π/3 and x = 2π/3
b) sin(x) = -√3 / 2:
In the third and fourth quadrants, sin(x) is negative.
Taking the inverse sine:
x = 4π/3 and x = 5π/3
Therefore, the solutions to the equation 2sin^2(x) + 3cos(x) = 3 in the interval [0, 2pi] are:
x = π/3, 2π/3, 4π/3, and 5π/3
Use the identity sin²(x)=1-cos²(x).
So
2sin^2x+3cosx=3 becomes
2cos²(x)-3cos(x)-1=0
Solve the quadratic for cos(x),
I get cos(x)=1 or cos(x)=1/2
On the interval [0,2π], each value of cos(x) has two solutions for a total of 4 values of x.