Question

find all angles, $0^{circ} leq x < 360^{circ}$, that satisfy the equation below, to the nearest 10th of a degree.

$4 cos x \tan x = 5 \tan x$

Explanation:

Step1: Rearrange the equation

First, we start with the given equation \(4\cos x\tan x = 5\tan x\). We can rearrange it by subtracting \(5\tan x\) from both sides to get \(4\cos x\tan x - 5\tan x = 0\). Then, we factor out \(\tan x\) from the left - hand side: \(\tan x(4\cos x - 5)=0\).

Step2: Solve for \(\tan x = 0\)

When \(\tan x=0\), we know that the tangent function \(y = \tan x\) has a value of \(0\) when \(x = k\cdot180^{\circ}\), where \(k\in\mathbb{Z}\). Since we are looking for solutions in the interval \(0^{\circ}\leq x<360^{\circ}\), when \(k = 0\), \(x = 0^{\circ}\); when \(k = 1\), \(x=180^{\circ}\).

Step3: Solve for \(4\cos x-5 = 0\)

We solve the equation \(4\cos x-5 = 0\) for \(\cos x\). First, we add \(5\) to both sides: \(4\cos x=5\), then divide both sides by \(4\) to get \(\cos x=\frac{5}{4}=1.25\). But we know that the range of the cosine function \(y = \cos x\) is \(- 1\leq\cos x\leq1\), and \(1.25>1\), so there are no solutions for this part of the equation.

Answer:

\(x = 0^{\circ},180^{\circ}\)