\( ( 5.4 ) \quad \cos 3 x = \cos ( x + 20 ^ { \circ } ) \)

To solve the equation \( \cos 3x = \cos (x + 20^\circ) \), we can use the sum-to-product identity for cosine: \[ \cos (a + b) = \cos a \cos b - \sin a \sin b \] Substitute \( a = x \) and \( b = 20^\circ \) into the equation: \[ \cos 3x = \cos x \cos 20^\circ - \sin x \sin 20^\circ \] Now, we can equate the two expressions: \[ \cos 3x = \cos x \cos 20^\circ - \sin x \sin 20^\circ \] This equation can be solved using trigonometric identities and properties. Let's simplify it further. Solve the equation \( \cos 3x = \cos x \cos 20^\circ - \sin x \sin 20^\circ \). Solve the equation by following steps: - step0: Solve for \(x\): \(\cos\left(3x\right)=\cos\left(x\right)\cos\left(20^{\circ}\right)-\sin\left(x\right)\sin\left(20^{\circ}\right)\) - step1: Transform the expression: \(\cos\left(3x\right)=\cos\left(x+20^{\circ}\right)\) - step2: Move the expression to the left side: \(\cos\left(3x\right)-\cos\left(x+20^{\circ}\right)=0\) - step3: Transform the expression: \(-2\sin\left(\frac{4x+20^{\circ}}{2}\right)\sin\left(\frac{2x-20^{\circ}}{2}\right)=0\) - step4: Elimination the left coefficient: \(\sin\left(\frac{4x+20^{\circ}}{2}\right)\sin\left(\frac{2x-20^{\circ}}{2}\right)=0\) - step5: Separate into possible cases: \(\begin{align}&\sin\left(\frac{4x+20^{\circ}}{2}\right)=0\\&\sin\left(\frac{2x-20^{\circ}}{2}\right)=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x=\frac{17\pi }{36}+\frac{k\pi }{2},k \in \mathbb{Z}\\&x=10^{\circ}+k\pi ,k \in \mathbb{Z}\end{align}\) - step7: Find the union: \(x=\left\{ \begin{array}{l}10^{\circ}+k\pi \\\frac{17\pi }{36}+\frac{k\pi }{2}\end{array}\right.,k \in \mathbb{Z}\) The solution to the equation \( \cos 3x = \cos (x + 20^\circ) \) is: \[ x = 10^\circ + k\pi \quad \text{or} \quad x = \frac{17\pi}{36} + \frac{k\pi}{2} \] where \( k \) is an integer.