Questions: Find all angles, (0^circ leq x<360^circ), that satisfy the equation below, to the nearest 10th of a degree. [2 cos x sin x-3 sin x=0]

Transcript text: Find all angles, $0^{\circ} \leq x<360^{\circ}$, that satisfy the equation below, to the nearest 10th of a degree. \[ 2 \cos x \sin x-3 \sin x=0 \]

Solution

Solution Steps

Step 1: Factor the Equation

We start with the equation: \[ 2 \cos x \sin x - 3 \sin x = 0 \] Factoring out $\sin x$, we get: \[ \sin x (2 \cos x - 3) = 0 \]

Step 2: Solve for $\sin x = 0$

Setting $\sin x = 0$, we find the angles where the sine function is zero: \[ x = 0^{\circ}, 180^{\circ} \]

Step 3: Solve for $2 \cos x - 3 = 0$

Next, we solve the equation: \[ 2 \cos x - 3 = 0 \] Rearranging gives: \[ \cos x = \frac{3}{2} \] Since the cosine function cannot exceed 1, there are no valid solutions from this equation.

Step 4: Compile All Solutions

Combining the solutions from $\sin x = 0$, we have: \[ x = 0^{\circ}, 180^{\circ} \] Additionally, since $x$ must satisfy $0^{\circ} \leq x < 360^{\circ}$, we exclude $360^{\circ}$ as it is not within the specified range.

Final Solutions

The angles that satisfy the original equation are: \[ x = 0^{\circ}, 180^{\circ} \]