Solve the equation on the interval [tex]$0 \leq \theta\ \textless \ 2 \pi$[/tex].
[tex]$\sin \left(2 \theta-\frac{\pi}{2}\right)=-1$[/tex]
What are the solutions in the interval [tex]$0 \leq \theta\ \textless \ 2 \pi$[/tex]?

The solutions to the equation sin ( 2 θ − 2 π ​ ) = − 1 in the interval 0 ≤ θ < 2 π are 0 and π .
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Answered by Anonymous | 2025-07-04

Substitute u = 2 θ − 2 π ​ and solve sin ( u ) = − 1 , which gives u = 2 3 π ​ + 2 nπ .
Substitute back to get 2 θ − 2 π ​ = 2 3 π ​ + 2 nπ .
Solve for θ to find θ = ( n + 1 ) π .
Find the solutions in the interval 0 ≤ θ < 2 π , which are 0 , π ​ .

Explanation

Problem Analysis We are given the equation sin ( 2 θ − 2 π ​ ) = − 1 and we want to find all solutions θ in the interval 0 ≤ θ < 2 π .

Substitution and General Solution Let u = 2 θ − 2 π ​ . Then the equation becomes sin ( u ) = − 1 . We know that sin ( u ) = − 1 when u = 2 3 π ​ + 2 nπ , where n is an integer.

Back-Substitution Substituting back u = 2 θ − 2 π ​ , we have 2 θ − 2 π ​ = 2 3 π ​ + 2 nπ .

Solving for Theta Now, we solve for θ : 2 θ = 2 3 π ​ + 2 π ​ + 2 nπ = 2 4 π ​ + 2 nπ = 2 π + 2 nπ Dividing by 2, we get: θ = π + nπ = ( n + 1 ) π

Finding Solutions in the Interval We want to find the solutions in the interval 0 ≤ θ < 2 π . We test different integer values of n .

Solution for n = -1 For n = − 1 , we have θ = ( − 1 + 1 ) π = 0 π = 0 . Since 0 ≤ 0 < 2 π , θ = 0 is a solution.

Solution for n = 0 For n = 0 , we have θ = ( 0 + 1 ) π = 1 π = π . Since 0 ≤ π < 2 π , θ = π is a solution.

Solution for n = 1 For n = 1 , we have θ = ( 1 + 1 ) π = 2 π . Since 0 ≤ 2 π < 2 π is false, θ = 2 π is not a solution.

Final Solutions For n = − 2 , we have θ = ( − 2 + 1 ) π = − 1 π = − π . Since 0 ≤ − π < 2 π is false, θ = − π is not a solution. Thus, the solutions in the interval 0 ≤ θ < 2 π are 0 and π .


Examples
Understanding trigonometric equations is crucial in many fields, such as physics and engineering. For example, when analyzing the motion of a pendulum, the angle θ it makes with the vertical direction can be modeled by a trigonometric function. Solving equations like the one above helps determine the specific times when the pendulum reaches certain positions. Similarly, in electrical engineering, alternating current (AC) circuits involve sinusoidal functions, and solving trigonometric equations is essential for calculating voltage and current values at different times.

Answered by GinnyAnswer | 2025-07-04