We are given two trigonometric equations. Let's simplify the first equation using the double angle identity for cosine, cos 2θ = 1 - 2sin² θ. Substituting this into the first equation will help us find the possible values of sin θ.

We are given two trigonometric equations. Let's simplify the first equation using the double angle identity for cosine, cos 2θ = 1 - 2sin² θ. Substituting this into the first equation will help us find the possible values of sin θ.

After substituting the identity, we rearrange the terms to solve for sin² θ. This gives us sin² θ = (1)/(4), which means sin θ can be either (1)/(2) or -(1)/(2).

Now, let's simplify the second equation. We can use the identity cos² θ = 1 - sin² θ to express the equation entirely in terms of sin θ. This will allow us to check which values of sin θ from the previous step satisfy this equation.

Now, let's simplify the second equation. We can use the identity cos² θ = 1 - sin² θ to express the equation entirely in terms of sin θ. This will allow us to check which values of sin θ from the previous step satisfy this equation.

We can factor the quadratic equation 2sin² θ + 3sin θ - 2 = 0 into (2sin θ - 1)(sin θ + 2) = 0. This gives two possible values for sin θ: (1)/(2) or -2. Since the range of sin θ is [-1, 1], sin θ = -2 is not a valid solution.

We can factor the quadratic equation 2sin² θ + 3sin θ - 2 = 0 into (2sin θ - 1)(sin θ + 2) = 0. This gives two possible values for sin θ: (1)/(2) or -2. Since the range of sin θ is [-1, 1], sin θ = -2 is not a valid solution.

Comparing the solutions from both equations, the only common value for sin θ is (1)/(2). We need to find all values of θ in the interval [0, 2π] for which sin θ = (1)/(2).

In the interval [0, 2π], the values of θ for which sin θ = (1)/(2) are (π)/(6) (in the first quadrant) and (5π)/(6) (in the second quadrant). The sum of these values is (π)/(6) + (5π)/(6) = (6π)/(6) = π.

In the interval [0, 2π], the values of θ for which sin θ = (1)/(2) are (π)/(6) (in the first quadrant) and (5π)/(6) (in the second quadrant). The sum of these values is (π)/(6) + (5π)/(6) = (6π)/(6) = π.

The sum of all values of $\theta \in [0, 2\pi]$ satisfying $2\sin^2 \theta = \cos 2\theta$ and $2\cos^2 \theta = 3\sin \theta$ is:

Answer: $\pi$

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, $\cos 2\theta = 1 - 2\sin^2 \theta$ . Substituting this into the first equation will help us find the possible values of $\sin \theta$ .

Step 2: Solve for $\sin \theta$ from the first equation

After substituting the identity, we rearrange the terms to solve for $\sin^2 \theta$ . This gives us $\sin^2 \theta = \frac{1}{4}$ , which means $\sin \theta$ can be either $\frac{1}{2}$ or $-\frac{1}{2}$ .

Step 3: Simplify the second equation

Now, let's simplify the second equation. We can use the identity $\cos^2 \theta = 1 - \sin^2 \theta$ to express the equation entirely in terms of $\sin \theta$ . This will allow us to check which values of $\sin \theta$ from the previous step satisfy this equation.

Step 4: Solve for $\sin \theta$ from the second equation

Substitute $\cos^2 \theta = 1 - \sin^2 \theta$ into the second equation. This results in a quadratic equation in terms of $\sin \theta$ . Let $x = \sin \theta$ , so the equation becomes $2x^2 + 3x - 2 = 0$ .

Step 5: Solve the quadratic equation for $\sin \theta$

We can factor the quadratic equation $2\sin^2 \theta + 3\sin \theta - 2 = 0$ into $(2\sin \theta - 1)(\sin \theta + 2) = 0$ . This gives two possible values for $\sin \theta$ : $\frac{1}{2}$ or $-2$ . Since the range of $\sin \theta$ is $[-1, 1]$ , $\sin \theta = -2$ is not a valid solution.

Step 6: Find common solutions for $\sin \theta$

Comparing the solutions from both equations, the only common value for $\sin \theta$ is $\frac{1}{2}$ . We need to find all values of $\theta$ in the interval $[0, 2\pi]$ for which $\sin \theta = \frac{1}{2}$ .

Step 7: Find $\theta$ values and their sum

In the interval $[0, 2\pi]$ , the values of $\theta$ for which $\sin \theta = \frac{1}{2}$ are $\frac{\pi}{6}$ (in the first quadrant) and $\frac{5\pi}{6}$ (in the second quadrant). The sum of these values is $\frac{\pi}{6} + \frac{5\pi}{6} = \frac{6\pi}{6} = \pi$ .

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The sum of all values of θ∈[0,2π]\theta \in [0, 2\pi]θ∈[0,2π] satisfying 2sin⁡2θ=cos⁡2θ2\sin^2 \theta = \cos 2\theta2sin2θ=cos2θ and 2cos⁡2θ=3sin⁡θ2\cos^2 \theta = 3\sin \theta2cos2θ=3sinθ is:

Step-by-step solution

Step 1: Simplify the first equation

Step 2: Solve for sin⁡θ\sin \thetasinθ from the first equation

Step 3: Simplify the second equation

Step 4: Solve for sin⁡θ\sin \thetasinθ from the second equation

Step 5: Solve the quadratic equation for sin⁡θ\sin \thetasinθ

Step 6: Find common solutions for sin⁡θ\sin \thetasinθ

Step 7: Find θ\thetaθ values and their sum