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

We wish to solve the following equations for all solutions: (a) cos 2θ = 0 (b) 5 sin θ + 14 cos θ = 0 ───────────────────────────── Solution (a): Step 1. Recognize that cos φ = 0 when   φ = π/2 + πk,  k ∈ ℤ. Step 2. Here, set φ = 2θ. Then the equation cos 2θ = 0 becomes:   2θ = π/2 + πk,  k ∈ ℤ. Step 3. Solve for θ by dividing both sides by 2:   θ = (π/4) + (π/2)k,  k ∈ ℤ. Thus, the general solution for (a) is:   θ = π/4 + (π/2)k,  k ∈ ℤ. ───────────────────────────── Solution (b): We have the equation:   5 sin θ + 14 cos θ = 0 Step 1. Assuming cos θ ≠ 0 (we will check for the extraneous possibility later), divide by cos θ:   5 (sin θ/cos θ) + 14 = 0   => 5 tan θ + 14 = 0 Step 2. Solve for tan θ:   tan θ = -14/5. Step 3. The general solution for tan θ = -14/5 is:   θ = arctan(–14/5) + πk,  k ∈ ℤ. Note: Since arctan(–14/5) is a particular solution, this form covers all solutions of the equation. Thus, the general solution for (b) is:   θ = arctan(–14/5) + πk,  k ∈ ℤ. ───────────────────────────── Final Answers: (a) θ = π/4 + (π/2)k,  k ∈ ℤ (b) θ = arctan(–14/5) + πk,  k ∈ ℤ