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]
Answer (1)
Substitute u = 2 θ − 2 π , transforming the equation to sin ( u ) = − 1 .
Solve for u , obtaining u = 2 3 π + 2 nπ , where n is an integer.
Substitute back to find θ = ( n + 1 ) π .
Determine the values of n for which 0 ≤ θ < 2 π , yielding the solutions 0 , π .
Explanation
Understanding the Problem We are asked to solve the equation sin ( 2 θ − 2 π ) = − 1 on the interval 0 ≤ θ < 2 π . This means we need to find all values of θ within this interval that satisfy the given equation.
Making a Substitution Let's make a substitution to simplify the equation. Let u = 2 θ − 2 π . Then the equation becomes sin ( u ) = − 1 .
Finding the General Solution for u Now we need to find the general solution for u . We know that sin ( u ) = − 1 when u = 2 3 π + 2 nπ , where n is an integer.
Substituting Back Substitute back u = 2 θ − 2 π to get 2 θ − 2 π = 2 3 π + 2 nπ .
Solving for theta Solve for θ : 2 θ = 2 3 π + 2 π + 2 nπ = 2 π + 2 nπ . Thus, θ = π + nπ = ( n + 1 ) π .
Finding Solutions in the Interval Now we need to find the values of n such that 0 ≤ θ < 2 π . We can test different integer values of n to find the solutions within the given interval.
For n = − 1 , θ = ( − 1 + 1 ) π = 0 π = 0 .
For n = 0 , θ = ( 0 + 1 ) π = 1 π = π .
For n = 1 , θ = ( 1 + 1 ) π = 2 π . However, 2 π is not in the interval [ 0 , 2 π ) , so it's not a solution. For n = − 2 , θ = ( − 2 + 1 ) π = − 1 π = − π . However, − π is not in the interval [ 0 , 2 π ) , so it's not a solution.
Final Answer Therefore, the solutions are θ = 0 and θ = π .
Examples
Understanding trigonometric equations is crucial in various fields like physics and engineering. For instance, when analyzing the motion of a pendulum, the angle it makes with the vertical can be modeled using trigonometric functions. Solving equations like the one above helps determine 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 points in time. These applications highlight the practical importance of mastering trigonometric equations.
