Find all of the solutions between 0 and 2pi:
2sin(x)^2 = 2 + cos(x)
I assume you meant (sin(X))^2
change the sin squared to 1-cos squared, then group the terms as in a quadratic, and solve with the quadratic equation.
To find the solutions to the equation 2sin(x)^2 = 2 + cos(x) between 0 and 2pi, we can follow these steps:
Step 1: Simplify the equation:
Start by simplifying the left side of the equation using the identity sin^2(x) + cos^2(x) = 1.
Since sin^2(x) = 1 - cos^2(x), we can rewrite the equation as:
2(1-cos^2(x)) = 2 + cos(x)
Expanding the equation gives:
2 - 2cos^2(x) = 2 + cos(x)
Rearranging the terms:
2cos^2(x) + cos(x) = 0
Step 2: Factor the equation:
To solve this equation, we need to factor it. Looking at the equation, we can see that it is a quadratic equation in terms of cos(x). Factor out cos(x) from the equation:
cos(x)(2cos(x) + 1) = 0
Using the zero product property, we set each factor equal to zero:
cos(x) = 0
2cos(x) + 1 = 0
Step 3: Solve each equation:
For the equation cos(x) = 0, we need to find the values of x between 0 and 2pi where the cosine function equals zero. These values occur at x = pi/2 and x = 3pi/2.
For the equation 2cos(x) + 1 = 0, we can solve for cos(x) by subtracting 1 from both sides and dividing by 2:
2cos(x) = -1
cos(x) = -1/2
The values of x between 0 and 2pi where cos(x) equals -1/2 occur at x = 2pi/3 and x = 4pi/3.
Step 4: Combine the solutions:
The solutions to the original equation 2sin(x)^2 = 2 + cos(x) between 0 and 2pi are:
x = pi/2, 2pi/3, 3pi/2, and 4pi/3.