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Solve the following trigonometric equations, finding all real solutions in the interval [0,2 π) and all

general solutions. Your work must support your answers.
2sin θ =1

User Mugentoki
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1 Answer

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Final answer:

To solve the equation 2sin(θ) = 1 in the interval [0, 2π), divide both sides by 2 to get sin(θ) = 1/2. The solutions are θ = π/6 + 2πk and θ = 5π/6 + 2πk, where k is an integer.

Step-by-step explanation:

To solve the equation 2sin(θ) = 1 in the interval [0, 2π), we can start by dividing both sides of the equation by 2, which gives us sin(θ) = 1/2. This equation tells us that the sine of an angle θ is equal to 1/2.

In order to find the values of θ that satisfy this equation, we can look at the values of θ for which sin(θ) = 1/2. From the unit circle or trigonometric identities, we know that the angles at which sin(θ) = 1/2 are π/6 and 5π/6.

Since we are looking for solutions in the interval [0, 2π), we can add any multiple of 2π to our solutions to find the general solutions. Therefore, the solutions to the equation 2sin(θ) = 1 in the interval [0, 2π) are θ = π/6 + 2πk and θ = 5π/6 + 2πk, where k is an integer.

User Karthik Nayak
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