Let [tex]$\frac{\cos (2 x)}{\cos (x)+\sin (x)}=0$[/tex] where [tex]$0^{\circ} \leq x \leq 180^{\circ}$[/tex]. What are the possible values for [tex]$x$[/tex]?
A. [tex]$45^{\circ}$[/tex] only
B. [tex]$135^{\circ}$[/tex] only
C. [tex]$45^{\circ}$[/tex] or [tex]$225^{\circ}$[/tex]
D. [tex]$135^{\circ}$[/tex] or [tex]$315^{\circ}$[/tex]
Answer (2)
The equation is zero when the numerator is zero and the denominator is not zero.
Solve cos ( 2 x ) = 0 , which gives x = 4 5 ∘ + 9 0 ∘ k .
Find the values of x in the interval 0 ∘ ≤ x ≤ 18 0 ∘ , which are 4 5 ∘ and 13 5 ∘ .
Check if cos ( x ) + sin ( x ) = 0 for these values. x = 4 5 ∘ satisfies this condition, but x = 13 5 ∘ does not.
The only possible value for x is 4 5 ∘ .
Explanation
Problem Analysis We are given the equation c o s ( x ) + s i n ( x ) c o s ( 2 x ) = 0 and we need to find the possible values of x in the interval 0 ∘ ≤ x ≤ 18 0 ∘ .
Solving Strategy A fraction is equal to zero if and only if its numerator is zero and its denominator is not zero. Therefore, we need to solve cos ( 2 x ) = 0 and ensure that cos ( x ) + sin ( x ) = 0 .
Solving the Numerator First, let's solve cos ( 2 x ) = 0 . The general solution for cos ( θ ) = 0 is θ = 9 0 ∘ + 18 0 ∘ k , where k is an integer. Thus, 2 x = 9 0 ∘ + 18 0 ∘ k , which gives x = 4 5 ∘ + 9 0 ∘ k .
Finding Possible Solutions Now, we find the values of x in the interval 0 ∘ ≤ x ≤ 18 0 ∘ .
For k = 0 , x = 4 5 ∘ .
For k = 1 , x = 4 5 ∘ + 9 0 ∘ = 13 5 ∘ .
For k = 2 , x = 4 5 ∘ + 18 0 ∘ = 22 5 ∘ , which is outside the given interval. For k = − 1 , x = 4 5 ∘ − 9 0 ∘ = − 4 5 ∘ , which is outside the given interval. So, the possible values of x are 4 5 ∘ and 13 5 ∘ .
Checking the Denominator Next, we need to check if cos ( x ) + sin ( x ) = 0 for these values. For x = 4 5 ∘ , cos ( 4 5 ∘ ) + sin ( 4 5 ∘ ) = 2 2 + 2 2 = 2 = 0 .
For x = 13 5 ∘ , cos ( 13 5 ∘ ) + sin ( 13 5 ∘ ) = − 2 2 + 2 2 = 0 . Therefore, x = 13 5 ∘ is not a valid solution.
Final Answer Therefore, the only possible value for x is 4 5 ∘ .
Examples
Understanding trigonometric equations is crucial in various fields such as physics and engineering. For instance, when analyzing the motion of a pendulum or the behavior of alternating current in an electrical circuit, solving equations involving trigonometric functions like sine and cosine is essential to determine key parameters such as amplitude, frequency, and phase. These parameters help in designing and optimizing systems that rely on oscillatory behavior.
To solve the equation, we find that cos ( 2 x ) = 0 gives solutions of 4 5 ∘ and 13 5 ∘ . However, only 4 5 ∘ is valid since it keeps the denominator non-zero. Thus, the answer is 4 5 ∘ .
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