Solve: [tex]$\sin \left(\frac{11 \pi}{2}+x\right)=-\frac{1}{2}$[/tex] for [tex]$[\pi, 2 \pi]$[/tex]
Answer (2)
Simplify the equation using trigonometric identities: sin ( 2 11 π + x ) = sin ( 2 3 π + x ) = − cos ( x ) .
Solve the simplified equation: − cos ( x ) = − 2 1 ⟹ cos ( x ) = 2 1 .
Find the general solutions: x = ± 3 π + 2 πk , where k is an integer.
Identify the solution in the interval [ π , 2 π ] : x = 3 5 π .
3 5 π
Explanation
Problem Analysis We are given the equation sin ( 2 11 π + x ) = − 2 1 and we need to solve for x in the interval [ π , 2 π ] .
Simplifying the Argument First, let's simplify the argument of the sine function. Since the sine function has a period of 2 π , we can subtract multiples of 2 π from 2 11 π without changing the value of the sine function. We have 2 11 π = 2 3 π + 4 π , so sin ( 2 11 π + x ) = sin ( 2 3 π + x ) .
Applying Sine Addition Formula Thus, our equation becomes sin ( 2 3 π + x ) = − 2 1 . Using the sine addition formula, we have sin ( 2 3 π + x ) = sin ( 2 3 π ) cos ( x ) + cos ( 2 3 π ) sin ( x ) = ( − 1 ) cos ( x ) + ( 0 ) sin ( x ) = − cos ( x ) .
Simplifying the Equation So, our equation is − cos ( x ) = − 2 1 , which simplifies to cos ( x ) = 2 1 .
Finding General Solutions Now we need to find the values of x in the interval [ π , 2 π ] such that cos ( x ) = 2 1 . We know that cos ( 3 π ) = 2 1 . Since cosine is positive in the first and fourth quadrants, the solutions to cos ( x ) = 2 1 are of the form x = 3 π + 2 πk and x = − 3 π + 2 πk , where k is an integer.
Finding Solutions in the Interval We want to find the solutions in the interval [ π , 2 π ] . For x = 3 π + 2 πk , if k = 0 , then x = 3 π , which is not in the interval. If k = 1 , then x = 3 π + 2 π = 3 7 π , which is greater than 2 π , so it's not in the interval. For x = − 3 π + 2 πk , if k = 0 , then x = − 3 π , which is not in the interval. If k = 1 , then x = − 3 π + 2 π = 3 5 π . Since π < 3 5 π < 2 π , this solution is in the interval. If k = 2 , then x = − 3 π + 4 π = 3 11 π , which is greater than 2 π , so it's not in the interval.
Final Solution Therefore, the only solution in the interval [ π , 2 π ] is x = 3 5 π .
Examples
Trigonometric equations are used in physics to model oscillations and waves. For example, the height of a wave can be modeled using a sine function. Solving trigonometric equations allows us to find the times at which the wave reaches a certain height. This is useful in many applications, such as designing bridges and buildings that can withstand the forces of nature.
The solution to the equation sin ( 2 11 π + x ) = − 2 1 for x in the interval [ π , 2 π ] is x = 3 5 π . This result was obtained by simplifying the sine argument and solving for cos ( x ) .
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