Math / Trigonometry

Questionc $3 \sin 2 x-\cos 2 x=0$

Studdy Solution

STEP 1

1. The equation $3 \sin 2x - \cos 2x = 0$ is a trigonometric equation.
2. We will solve for $x$ in terms of angles that satisfy the equation.
3. The solutions will be within the standard interval for trigonometric functions, typically $0 \leq x < 2\pi$ .

STEP 2

1. Isolate one trigonometric function.
2. Use trigonometric identities to simplify.
3. Solve for $x$ within the specified interval.

STEP 3

First, isolate one of the trigonometric functions. Let's isolate $\sin 2x$ :
$3 \sin 2x - \cos 2x = 0$ $3 \sin 2x = \cos 2x$

STEP 4

Divide both sides by $\cos 2x$ to express $\sin 2x$ in terms of $\tan 2x$ :
$\frac{3 \sin 2x}{\cos 2x} = 1$ $3 \tan 2x = 1$

STEP 5

Solve for $\tan 2x$ :
$\tan 2x = \frac{1}{3}$
Now, solve for $2x$ using the inverse tangent function:
$2x = \tan^{-1}\left(\frac{1}{3}\right) + n\pi$
where $n$ is an integer because the tangent function has a period of $\pi$ .

STEP 6

Solve for $x$ :
$x = \frac{1}{2}\left(\tan^{-1}\left(\frac{1}{3}\right) + n\pi\right)$
Determine the values of $x$ within the interval $0 \leq x < 2\pi$ :
1. For $n = 0$ : $x = \frac{1}{2}\tan^{-1}\left(\frac{1}{3}\right)$
2. For $n = 1$ : $x = \frac{1}{2}\left(\tan^{-1}\left(\frac{1}{3}\right) + \pi\right)$
Continue this process to find all solutions within the interval.
The solutions for $x$ are:
$x = \frac{1}{2}\tan^{-1}\left(\frac{1}{3}\right) + \frac{n\pi}{2}$
where $n$ is an integer such that $0 \leq x < 2\pi$ .

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