Find all solutions on the interval [0,2pi] for the following:
2sin^2(x)-5sin(x)=-3
cos^2(x)+sin(x)=1
To find all solutions on the interval [0, 2pi] for these equations, we can use algebraic manipulations and basic trigonometric identities. Let's solve each equation step-by-step:
1. Equation: 2sin^2(x) - 5sin(x) = -3
Step 1: Rearrange the equation to have a quadratic expression.
2sin^2(x) - 5sin(x) + 3 = 0
Step 2: Factor the quadratic expression.
(2sin(x) - 3)(sin(x) - 1) = 0
Step 3: Solve the two separate equations.
1st Equation: 2sin(x) - 3 = 0
sin(x) = 3/2
Since sin(x) cannot be greater than 1, there are no solutions for this equation.
2nd Equation: sin(x) - 1 = 0
sin(x) = 1
x = pi/2
Therefore, the solution for the equation 2sin^2(x) - 5sin(x) = -3 on the interval [0, 2pi] is x = pi/2.
2. Equation: cos^2(x) + sin(x) = 1
Step 1: Apply the Pythagorean identity cos^2(x) + sin^2(x) = 1 to the equation.
cos^2(x) = 1 - sin(x)
Step 2: Substitute the value of cos^2(x) in the equation.
1 - sin^2(x) + sin(x) = 1
Step 3: Rearrange the equation.
-sin^2(x) + sin(x) = 0
Step 4: Factor out sin(x).
sin(x)(-sin(x) + 1) = 0
Step 5: Solve the two separate equations.
1st Equation: sin(x) = 0
x = 0, pi
2nd Equation: -sin(x) + 1 = 0
sin(x) = 1
x = pi/2
Therefore, the solutions for the equation cos^2(x) + sin(x) = 1 on the interval [0, 2pi] are x = 0, pi/2, and pi.