The sum of all values of θ∈[0,2π] satisfying 2sin2θ=cos2θ and 2cos2θ=3sinθ is:
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Step-by-Step Solution
Step 1: Simplify the first equation
We are given two trigonometric equations. Let's simplify the first equation using the double angle identity for cosine, cos2θ=1−2sin2θ. Substituting this into the first equation will help us find the possible values of sinθ.
Step 2: Solve for sinθ from the first equation
After substituting the identity, we rearrange the terms to solve for sin2θ. This gives us sin2θ=41, which means sinθ can be either 21 or −21.
Step 3: Simplify the second equation
Now, let's simplify the second equation. We can use the identity cos2θ=1−sin2θ to express the equation entirely in terms of sinθ. This will allow us to check which values of sinθ from the previous step satisfy this equation.
Step 4: Solve for sinθ from the second equation
Substitute cos2θ=1−sin2θ into the second equation. This results in a quadratic equation in terms of sinθ. Let x=sinθ, so the equation becomes 2x2+3x−2=0.
Step 5: Solve the quadratic equation for sinθ
We can factor the quadratic equation 2sin2θ+3sinθ−2=0 into (2sinθ−1)(sinθ+2)=0. This gives two possible values for sinθ: 21 or −2. Since the range of sinθ is [−1,1], sinθ=−2 is not a valid solution.
Step 6: Find common solutions for sinθ
Comparing the solutions from both equations, the only common value for sinθ is 21. We need to find all values of θ in the interval [0,2π] for which sinθ=21.
Step 7: Find θ values and their sum
In the interval [0,2π], the values of θ for which sinθ=21 are 6π (in the first quadrant) and 65π (in the second quadrant). The sum of these values is 6π+65π=66π=π.