General Solutions of Trigonometric Equations

 

For \( k \in \mathbb{Z} \):

\[ \cos f(x) = \cos g(x) \Rightarrow f(x) = g(x) + 2k\pi \quad \text{or} \quad f(x) = -g(x) + 2k\pi \]

\[ \sin f(x) = \sin g(x) \Rightarrow f(x) = g(x) + 2k\pi \quad \text{or} \quad f(x) = \pi – g(x) + 2k\pi \]

\[ \left. \begin{array}{l} \tan f(x) = \tan g(x) \\ \cot f(x) = \cot g(x) \end{array} \right\} \Rightarrow f(x) = g(x) + k\pi \]

 

Example:

 

Let’s find the solution set for the equation \( \cos \left(2x – \frac{\pi}{3}\right) = \cos \left(\frac{\pi}{4} – x\right) \).

\[ \cos \left(2x – \frac{\pi}{3}\right) = \cos \left(\frac{\pi}{4} – x\right) \]

**Branch 1:**
\[ 2x – \frac{\pi}{3} = \frac{\pi}{4} – x + 2k\pi \]
\[ 3x = \frac{\pi}{4} + \frac{\pi}{3} + 2k\pi \Rightarrow 3x = \frac{7\pi}{12} + 2k\pi \]
\[ x = \frac{7\pi}{36} + \frac{2k\pi}{3} = \frac{7\pi + 24k\pi}{36} \]

**Branch 2:**
\[ 2x – \frac{\pi}{3} = -\left(\frac{\pi}{4} – x\right) + 2k\pi \]
\[ 2x – \frac{\pi}{3} = -\frac{\pi}{4} + x + 2k\pi \]
\[ x = -\frac{\pi}{4} + \frac{\pi}{3} + 2k\pi \]
\[ x = \frac{\pi}{12} + 2k\pi = \frac{\pi + 24k\pi}{12} \quad (k \in \mathbb{Z}) \]

Therefore, the complete solution set for \( k \in \mathbb{Z} \) is:
\[ S = \left\{ x \;\middle|\; x = \frac{7\pi + 24k\pi}{36} \quad \text{or} \quad x = \frac{\pi + 24k\pi}{12} \right\} \]

 

Example:

 

Let’s find the solution set for the equation \( \sin \left(\frac{\pi}{3} – x\right) + \cos x = 0 \).

\[ \sin \left(\frac{\pi}{3} – x\right) + \cos x = 0 \]
\[ \Rightarrow \cos x = -\sin \left(\frac{\pi}{3} – x\right) \]
\[ \Rightarrow \cos x = \sin \left(-\frac{\pi}{3} + x\right) \]

Using the co-function identity \( \cos x = \sin \left(\frac{\pi}{2} – x\right) \):
\[ \Rightarrow \sin \left(\frac{\pi}{2} – x\right) = \sin \left(-\frac{\pi}{3} + x\right) \]

**Branch 1:**
\[ -\frac{\pi}{3} + x = \frac{\pi}{2} – x + 2k\pi \]
\[ 2x = \frac{\pi}{2} + \frac{\pi}{3} + 2k\pi \]
\[ 2x = \frac{5\pi}{6} + 2k\pi \Rightarrow x = \frac{5\pi}{12} + k\pi \]

**Branch 2:**
\[ -\frac{\pi}{3} + x = \pi – \left(\frac{\pi}{2} – x\right) + 2k\pi \]
\[ -\frac{\pi}{3} + x = \pi – \frac{\pi}{2} + x + 2k\pi \]
\[ -\frac{\pi}{3} \neq \frac{\pi}{2} + 2k\pi \quad \text{(No solution from this branch)} \]

Therefore, the solution set is:
\[ S = \left\{ x \;\middle|\; x = \frac{5\pi}{12} + k\pi, \ k \in \mathbb{Z} \right\} \]

 

Example:

 

Let’s find the solution set for the equation \( \tan 5x \cdot \tan 3x = 1 \).

\[ \tan 5x \cdot \tan 3x = 1 \Rightarrow \tan 5x = \frac{1}{\tan 3x} \]
\[ \Rightarrow \tan 5x = \cot 3x \]
\[ \Rightarrow \tan 5x = \tan \left(\frac{\pi}{2} – 3x\right) \]
\[ \Rightarrow 5x = \frac{\pi}{2} – 3x + k\pi \]
\[ \Rightarrow 8x = \frac{\pi}{2} + k\pi \]
\[ x = \frac{\pi}{16} + \frac{k\pi}{8} \quad (k \in \mathbb{Z}) \]

Therefore, the solution set is:
\[ S = \left\{ x \;\middle|\; x = \frac{\pi}{16} + \frac{k\pi}{8}, \ k \in \mathbb{Z} \right\} \]

 

QUESTION 64

 

What is the smallest positive value of $x$ that satisfies the equation \( \cos \left(x + \frac{\pi}{4}\right) + \sin \left(\frac{\pi}{4} – x\right) = 2 \cos \left(x + \frac{2\pi}{3}\right) \)?

\[ A) \frac{13\pi}{24} \quad B) \frac{7\pi}{12} \quad C) \frac{5\pi}{8} \quad D) \frac{3\pi}{4} \quad E) \frac{17\pi}{24} \]

 

Solution:

 

Using the identity \( \sin \theta = \cos \left(\frac{\pi}{2} – \theta\right) \), rewrite the sine term:
\[ \sin \left(\frac{\pi}{4} – x\right) = \cos \left[\frac{\pi}{2} – \left(\frac{\pi}{4} – x\right)\right] = \cos \left(\frac{\pi}{4} + x\right) \]

Substitute this back into the equation:
\[ \cos \left(x + \frac{\pi}{4}\right) + \cos \left(\frac{\pi}{4} + x\right) = 2 \cos \left(x + \frac{2\pi}{3}\right) \]
\[ 2 \cos \left(x + \frac{\pi}{4}\right) = 2 \cos \left(x + \frac{2\pi}{3}\right) \]
\[ \Rightarrow \cos \left(x + \frac{\pi}{4}\right) = \cos \left(x + \frac{2\pi}{3}\right) \]

**Branch 1:**
\[ x + \frac{\pi}{4} = x + \frac{2\pi}{3} + 2k\pi \Rightarrow \frac{\pi}{4} \neq \frac{2\pi}{3} + 2k\pi \quad \text{(No solution)} \]

**Branch 2:**
\[ x + \frac{\pi}{4} = -\left(x + \frac{2\pi}{3}\right) + 2k\pi \]
\[ 2x = -\frac{\pi}{4} – \frac{2\pi}{3} + 2k\pi \]
\[ 2x = -\frac{11\pi}{12} + 2k\pi \Rightarrow x = -\frac{11\pi}{24} + k\pi \]

To find the smallest positive root, set \( k = 1 \):
\[ x = -\frac{11\pi}{24} + \pi = \frac{13\pi}{24} \]

\( \textbf{Correct Answer: A} \)

 

QUESTION 65

 

What is the smallest positive value of $x$ that satisfies the equation \( \frac{\cos x}{\sin 7x} = \cot 3x \)?

\[ A) \frac{\pi}{16} \quad B) \frac{\pi}{12} \quad C) \frac{\pi}{10} \quad D) \frac{\pi}{8} \quad E) \frac{\pi}{6} \]

 

Solution:

 

Rewrite the cotangent term:
\[ \frac{\cos x}{\sin 7x} = \frac{\cos 3x}{\sin 3x} \Rightarrow \sin 3x \cdot \cos x = \sin 7x \cdot \cos 3x \]

Apply the product-to-sum identities \( \sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)] \):
\[ \frac{1}{2} (\sin 4x + \sin 2x) = \frac{1}{2} (\sin 10x + \sin 4x) \]
\[ \Rightarrow \sin 2x = \sin 10x \]

**Branch 1:**
\[ 10x = 2x + 2k\pi \Rightarrow 8x = 2k\pi \Rightarrow x = \frac{k\pi}{4} \]

**Branch 2:**
\[ 10x = \pi – 2x + 2k\pi \Rightarrow 12x = (2k + 1)\pi \Rightarrow x = \frac{(2k + 1)\pi}{12} \]

* For \( k = 0 \) in Branch 1: \( x = 0 \) (not positive)
* For \( k = 0 \) in Branch 2: \( x = \frac{\pi}{12} \)

Thus, the smallest positive value is \( \frac{\pi}{12} \).

\( \textbf{Correct Answer: B} \)

 

QUESTION 66

 

What is the sum of the roots of the equation \( \cos^2 x – \cos^2 \left(x – \frac{\pi}{3}\right) = 0 \) within the interval \( [0, \pi) \)?

\[ A) \frac{\pi}{6} \quad B) \frac{\pi}{3} \quad C) \frac{2\pi}{3} \quad D) \frac{4\pi}{3} \quad E) \frac{5\pi}{6} \]

 

Solution:

 

Factor using the difference of squares \( A^2 – B^2 = (A-B)(A+B) \):
\[ \left[\cos x – \cos \left(x – \frac{\pi}{3}\right)\right] \cdot \left[\cos x + \cos \left(x – \frac{\pi}{3}\right)\right] = 0 \]

Apply sum-to-product identities:
\[ \left[-2 \sin \left(x – \frac{\pi}{6}\right) \sin \left(\frac{\pi}{6}\right)\right] \cdot \left[2 \cos \left(x – \frac{\pi}{6}\right) \cos \left(\frac{\pi}{6}\right)\right] = 0 \]
\[ \Rightarrow -2 \sin \left(\frac{\pi}{6}\right) \cos \left(\frac{\pi}{6}\right) \cdot 2 \sin \left(x – \frac{\pi}{6}\right) \cos \left(x – \frac{\pi}{6}\right) = 0 \]

Using the double-angle formula \( 2\sin\theta\cos\theta = \sin(2\theta) \):
\[ \Rightarrow \sin \left(2x – \frac{\pi}{3}\right) = 0 \]
\[ \Rightarrow 2x – \frac{\pi}{3} = 0 + 2k\pi \quad \text{or} \quad 2x – \frac{\pi}{3} = \pi + 2k\pi \]
\[ \Rightarrow x = \frac{\pi}{6} + k\pi \quad \text{or} \quad x = \frac{2\pi}{3} + k\pi \]

Finding the roots inside \( [0, \pi) \) by setting \( k = 0 \):
\[ x_1 = \frac{\pi}{6}, \qquad x_2 = \frac{2\pi}{3} \]

Summing the valid roots:
\[ x_1 + x_2 = \frac{\pi}{6} + \frac{2\pi}{3} = \frac{5\pi}{6} \]

\( \textbf{Correct Answer: E} \)