solve sin(2θ)+1=0 on the interval 0≤θ≤2pi
To solve the equation sin(2θ) + 1 = 0 on the interval 0 ≤ θ ≤ 2π, we can follow these steps:
Step 1: Subtract 1 from both sides of the equation to isolate sin(2θ):
sin(2θ) = -1
Step 2: To solve for θ, we need to find the values of 2θ for which the sine function equals -1.
Step 3: Recall that the sine function is equal to -1 at specific angles. These angles are π/2 + 2πn, where n is an integer. In other words, sin(x) = -1 when x = π/2 + 2πn, where n is an integer.
Step 4: Since we have 2θ, we need to solve for θ by dividing both sides of the equation by 2:
2θ = π/2 + 2πn
Step 5: Divide both sides of the equation by 2 to solve for θ:
θ = (π/2 + 2πn)/2
Step 6: Simplify the expression to get the final answer for θ:
θ = π/4 + πn
Now, let's find the values of θ that satisfy 0 ≤ θ ≤ 2π.
For n = 0:
θ = π/4 + π * 0 = π/4
For n = 1:
θ = π/4 + π * 1 = 5π/4
Since the given interval is 0 ≤ θ ≤ 2π, the solutions for θ in the given equation sin(2θ) + 1 = 0 are θ = π/4 and θ = 5π/4.