7. Find the general solution of each of the following equations: \( \begin{array}{ll}\text { (a) } \cos 2 \theta=0,4 & \text { (b) } 5 \sin \theta+14 \cos \theta=0\end{array} \)

To find the general solution of each of the given equations, we will use trigonometric identities and properties. ### (a) \( \cos 2 \theta = 0.4 \) Given equation: \( \cos 2 \theta = 0.4 \) We know that \( \cos 2 \theta = 2 \cos^2 \theta - 1 \). Substituting this into the given equation, we get: \( 2 \cos^2 \theta - 1 = 0.4 \) Solving for \( \cos \theta \), we have: \( 2 \cos^2 \theta = 1.4 \) \( \cos^2 \theta = 0.7 \) \( \cos \theta = \pm \sqrt{0.7} \) Therefore, the general solution for \( \cos 2 \theta = 0.4 \) is: \( \cos \theta = \pm \sqrt{0.7} \) ### (b) \( 5 \sin \theta + 14 \cos \theta = 0 \) Given equation: \( 5 \sin \theta + 14 \cos \theta = 0 \) We can rewrite this equation in the form \( \sin \theta = -\frac{14}{5} \cos \theta \). Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), we have: \( \tan \theta = -\frac{14}{5} \) Therefore, the general solution for \( 5 \sin \theta + 14 \cos \theta = 0 \) is: \( \tan \theta = -\frac{14}{5} \)