Math / Trigonometry

QuestionDetermine the roots of the following equations for $-180^{\circ} \leq x \leq 180^{\circ}$ . b) $\sin 2 x-\cos 2 x=0$

Studdy Solution

STEP 1

What is this asking? We need to find the angles $x$ between $-180^\circ$ and $180^\circ$ that make $\sin 2x$ and $\cos 2x$ equal to each other. Watch out! The equation involves $2x$ , so we'll need to adjust our final answers to get the values for $x$ .
Also, remember to check for all possible solutions within the given range!

STEP 2

1. Rewrite the equation
2. Solve for $2x$
3. Find the general solutions for $2x$
4. Solve for $x$
5. Find specific solutions for $x$

STEP 3

Alright, let's rewrite our equation $\sin 2x - \cos 2x = 0$ by adding $\cos 2x$ to both sides.
This gives us $\sin 2x = \cos 2x$ .

STEP 4

Now, let's divide both sides by $\cos 2x$ to get $\frac{\sin 2x}{\cos 2x} = \frac{\cos 2x}{\cos 2x}$ , which simplifies to $\tan 2x = 1$ .
Awesome!

STEP 5

We know that $\tan 2x = 1$ .
The principal angle, the smallest positive angle where the tangent is 1, is $45^\circ$ .

STEP 6

Since the tangent function has a period of $180^\circ$ , the general solution for $2x$ is given by $2x = 45^\circ + 180^\circ n$ , where $n$ is any integer.

STEP 7

To find the general solution for $x$ , we divide our general solution for $2x$ by $2$ : $x = \frac{45^\circ + 180^\circ n}{2} = 22.5^\circ + 90^\circ n$ .

STEP 8

Now, let's find the specific solutions for $x$ in the range $-180^\circ \leq x \leq 180^\circ$ .
We'll do this by plugging in different integer values for $n$ into our general solution $x = 22.5^\circ + 90^\circ n$ .

STEP 9

When $n = 0$ , $x = 22.5^\circ + 90^\circ \cdot 0 = 22.5^\circ$ .
This is within our range!

STEP 10

When $n = 1$ , $x = 22.5^\circ + 90^\circ \cdot 1 = 112.5^\circ$ .
This is also within our range!

STEP 11

When $n = 2$ , $x = 22.5^\circ + 90^\circ \cdot 2 = 202.5^\circ$ .
This is outside our range, so we don't include it.

STEP 12

When $n = -1$ , $x = 22.5^\circ + 90^\circ \cdot (-1) = -67.5^\circ$ .
This is within our range!

STEP 13

When $n = -2$ , $x = 22.5^\circ + 90^\circ \cdot (-2) = -157.5^\circ$ .
This is within our range!

STEP 14

When $n = -3$ , $x = 22.5^\circ + 90^\circ \cdot (-3) = -247.5^\circ$ .
This is outside our range.

STEP 15

The solutions for $x$ in the given range are $x = -157.5^\circ$ , $-67.5^\circ$ , $22.5^\circ$ , and $112.5^\circ$ .

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