Linear Trigonometric Equations in Sine and Cosine
For \( a, b, c \in \mathbb{R} \), equations of the form
\[ \mathbf{a \sin x + b \cos x = c} \]
are called linear equations in \( \sin x \) and \( \cos x \). Different algebraic methods can be used to solve these equations.
Method 1: Using Auxiliary Angle Substitution
Divide both sides of the equation \( a \sin x + b \cos x = c \) by $a$:
\[ \sin x + \frac{b}{a} \cos x = \frac{c}{a} \]
Let \( \tan \alpha = \frac{b}{a} \). Substituting this gives:
\[ \Rightarrow \sin x + \tan \alpha \cdot \cos x = \frac{c}{a} \]
\[ \Rightarrow \sin x + \frac{\sin \alpha}{\cos \alpha} \cdot \cos x = \frac{c}{a} \]
Multiply through by \( \cos \alpha \):
\[ \Rightarrow \sin x \cos \alpha + \sin \alpha \cos x = \frac{c}{a} \cos \alpha \]
Using the sine sum identity \( \sin(A+B) = \sin A\cos B + \cos A\sin B \):
\[ \Rightarrow \sin (x + \alpha) = \frac{c}{a} \cos \alpha \]
Since \( \tan \alpha = \frac{b}{a} \), we can use a right triangle to find \( \cos \alpha = \frac{a}{\sqrt{a^2 + b^2}} \). Substituting this yields:
\[ \Rightarrow \sin (x + \alpha) = \frac{c}{\sqrt{a^2 + b^2}} \]
For this equation to have a real solution (\( S \neq \emptyset \)), the value on the right-hand side must fall within the range of the sine function:
\[ -1 \leq \frac{c}{\sqrt{a^2 + b^2}} \leq 1 \Rightarrow \left| \frac{c}{\sqrt{a^2 + b^2}} \right| \leq 1 \]
\[ \Rightarrow c^2 \leq a^2 + b^2 \]
Method 2: Using Weierstrass Half-Angle Substitution
Using the half-angle identities expressed in terms of \( \tan \left(\frac{x}{2}\right) \):
\[ \cos x = \frac{1 – \tan^2 \left(\frac{x}{2}\right)}{1 + \tan^2 \left(\frac{x}{2}\right)} \quad , \quad \sin x = \frac{2 \tan \left(\frac{x}{2}\right)}{1 + \tan^2 \left(\frac{x}{2}\right)} \]
Let \( t = \tan \left(\frac{x}{2}\right) \). Substitute these identities into \( a \sin x + b \cos x = c \):
\[ \Rightarrow a \cdot \frac{2t}{1 + t^2} + b \cdot \frac{1 – t^2}{1 + t^2} = c \]
Multiply the entire equation by \( (1 + t^2) \):
\[ \Rightarrow 2at + b (1 – t^2) = c (1 + t^2) \]
Rearrange into a standard quadratic equation form:
\[ \Rightarrow (b + c) t^2 – 2at – (b – c) = 0 \]
Solve this quadratic equation for $t$, then solve \( \tan \left(\frac{x}{2}\right) = t \) to find the values of $x$.
Example:
Find the solution set for the equation \( \sqrt{3} \sin x – 3 \cos x = \sqrt{3} \).
Divide both sides of the equation by \( \sqrt{3} \):
\[ \Rightarrow \sin x – \sqrt{3} \cos x = 1 \]
Substitute \( \sqrt{3} = \tan \left(\frac{\pi}{3}\right) \):
\[ \Rightarrow \sin x – \tan \left(\frac{\pi}{3}\right) \cdot \cos x = 1 \]
\[ \Rightarrow \sin x – \frac{\sin \left(\frac{\pi}{3}\right)}{\cos \left(\frac{\pi}{3}\right)} \cdot \cos x = 1 \]
Multiply through by \( \cos \left(\frac{\pi}{3}\right) \):
\[ \Rightarrow \sin x \cdot \cos \left(\frac{\pi}{3}\right) – \sin \left(\frac{\pi}{3}\right) \cdot \cos x = \cos \left(\frac{\pi}{3}\right) \]
Apply the sine difference identity:
\[ \Rightarrow \sin \left(x – \frac{\pi}{3}\right) = \cos \left(\frac{\pi}{3}\right) \]
Since \( \cos \left(\frac{\pi}{3}\right) = \frac{1}{2} = \sin \left(\frac{\pi}{6}\right) \):
\[ \Rightarrow \sin \left(x – \frac{\pi}{3}\right) = \sin \left(\frac{\pi}{6}\right) \]
This yields two branches of solutions for \( k \in \mathbb{Z} \):
\[ x – \frac{\pi}{3} = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x – \frac{\pi}{3} = \left(\pi – \frac{\pi}{6}\right) + 2k\pi \]
\[ x = \frac{\pi}{2} + 2k\pi \quad \text{or} \quad x = \frac{7\pi}{6} + 2k\pi \]
Therefore, the solution set is:
\[ S = \left\{ x \;\middle|\; x = \frac{\pi}{2} + 2k\pi \quad \text{or} \quad x = \frac{7\pi}{6} + 2k\pi, \ k \in \mathbb{Z} \right\} \]
Example:
Find the solution set for the equation \( \sin 2x – \cos 2x = \frac{\sqrt{2}}{2} \).
Rewrite the equation showing the implicit coefficient:
\[ \sin 2x – 1 \cdot \cos 2x = \frac{\sqrt{2}}{2} \]
Substitute \( 1 = \tan \left(\frac{\pi}{4}\right) \):
\[ \Rightarrow \sin 2x – \tan \left(\frac{\pi}{4}\right) \cdot \cos 2x = \frac{\sqrt{2}}{2} \]
\[ \Rightarrow \sin 2x – \frac{\sin \left(\frac{\pi}{4}\right)}{\cos \left(\frac{\pi}{4}\right)} \cdot \cos 2x = \frac{\sqrt{2}}{2} \]
Multiply through by \( \cos \left(\frac{\pi}{4}\right) \):
\[ \sin 2x \cdot \cos \left(\frac{\pi}{4}\right) – \sin \left(\frac{\pi}{4}\right) \cdot \cos 2x = \frac{\sqrt{2}}{2} \cdot \cos \left(\frac{\pi}{4}\right) \]
Apply the sine difference identity:
\[ \Rightarrow \sin \left(2x – \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{2} \]
Since \( \frac{1}{2} = \sin \left(\frac{\pi}{6}\right) \):
\[ \Rightarrow \sin \left(2x – \frac{\pi}{4}\right) = \sin \left(\frac{\pi}{6}\right) \]
This yields two branches of solutions for \( k \in \mathbb{Z} \):
\[ 2x – \frac{\pi}{4} = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad 2x – \frac{\pi}{4} = \frac{5\pi}{6} + 2k\pi \]
Isolating $x$:
\[ \Rightarrow x = \frac{5\pi}{24} + k\pi \quad \text{or} \quad x = \frac{13\pi}{24} + k\pi \]
Therefore, the solution set is:
\[ S = \left\{ x \;\middle|\; x = \frac{5\pi}{24} + k\pi \quad \text{or} \quad x = \frac{13\pi}{24} + k\pi, \ k \in \mathbb{Z} \right\} \]
Example:
Find the solution set for the equation \( 3 \sin x + 4 \cos x = 6 \).
Identify coefficients:
\[ a = 3, \ b = 4, \ c = 6 \Rightarrow a^2 + b^2 = 3 \dots 3^2 + 4^2 = 25 \quad \text{and} \quad c^2 = 6^2 = 36 \]
Check the existence condition:
\[ c^2 \leq a^2 + b^2 \Rightarrow 36 \leq 25 \quad \text{(False)} \]
Since the condition is not satisfied, the equation has no real solutions. The solution set is empty (\( \emptyset \)).