QuestionSolve for $x$ in the equation $2 \sin x = \sqrt{3} \tan x$ for $0^{\circ} \leq x \leq 360^{\circ}$ .

Studdy Solution

STEP 1

Assumptions1. The equation to solve is $\sin x=\sqrt{3} \tan x$ . The range of $x$ is $0^{\circ} \leq x \leq360^{\circ}$
3. We are looking for all possible values of $x$ within this range that satisfy the equation

STEP 2

We can rewrite the given equation in terms of sine and cosine to simplify it. $2 \sin x=\sqrt{} \tan x$ can be rewritten as $2 \sin x=\sqrt{} \frac{\sin x}{\cos x}$

STEP 3

Multiply both sides of the equation by $\cos x$ to eliminate the denominator on the right side.
$2 \sin x \cos x = \sqrt{3} \sin x$

STEP 4

Subtract $\sqrt{3} \sin x$ from both sides to set the equation to zero.
$2 \sin x \cos x - \sqrt{3} \sin x =0$

STEP 5

Factor out $\sin x$ from both sides.
$\sin x (2 \cos x - \sqrt{3}) =0$

STEP 6

Set each factor equal to zero and solve for $x$ .
$\sin x =0$ $2 \cos x - \sqrt{3} =0$

STEP 7

olve for $x$ in the first equation.
$x = \sin^{-1}(0)$

STEP 8

Find the values of $x$ that satisfy the first equation within the given range.
$x =0^{\circ},180^{\circ},360^{\circ}$

STEP 9

olve for $\cos x$ in the second equation.
$2 \cos x = \sqrt{3}$ $\cos x = \frac{\sqrt{3}}{2}$

STEP 10

olve for $x$ in the second equation.
$x = \cos^{-}\left(\frac{\sqrt{3}}{2}\right)$

STEP 11

Find the values of $x$ that satisfy the second equation within the given range.
$x =30^{\circ},330^{\circ}$ So, the values of $x$ that satisfy the equation $\sin x=\sqrt{3} \tan x$ within the range $0^{\circ} \leq x \leq360^{\circ}$ are $0^{\circ},30^{\circ},180^{\circ},330^{\circ},360^{\circ}$ .

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