Math / Trigonometry

Questionc) $7 \sec x=7$ f) $8+4 \cot x=10$
Using a calculator, determine the solutions for each equation, to two decimal places, on the interval $0 \leq x \leq 2 \pi$ . a) $\sin 2 x=\frac{1}{\sqrt{2}}$ c) $\sin 3 x=-\frac{\sqrt{3}}{2}$ e) $\cos 2 x=-\frac{1}{2}$

Studdy Solution

STEP 1

1. We are solving trigonometric equations within the interval $0 \leq x \leq 2\pi$ .
2. We will use inverse trigonometric functions and properties of trigonometric identities.
3. Calculators will be used to determine precise decimal values.

STEP 2

1. Solve equation $c) \, 7 \sec x = 7$ .
2. Solve equation $f) \, 8 + 4 \cot x = 10$ .
3. Solve equation $a) \, \sin 2x = \frac{1}{\sqrt{2}}$ .
4. Solve equation $c) \, \sin 3x = -\frac{\sqrt{3}}{2}$ .
5. Solve equation $e) \, \cos 2x = -\frac{1}{2}$ .

STEP 3

For $7 \sec x = 7$ , divide both sides by 7:
$\sec x = 1$
The secant function is the reciprocal of the cosine function, so:
$\cos x = 1$
The cosine of $x$ is 1 at $x = 0$ within the interval $0 \leq x \leq 2\pi$ .
Solution for $c)$ : $x = 0$ .

STEP 4

For $8 + 4 \cot x = 10$ , subtract 8 from both sides:
$4 \cot x = 2$
Divide by 4:
$\cot x = \frac{1}{2}$
The cotangent function is the reciprocal of the tangent function, so:
$\tan x = 2$
Use a calculator to find $x$ :
$x = \tan^{-1}(2) \approx 1.11$
Since the tangent function is periodic with period $\pi$ , the other solution in the interval is:
$x = \pi + 1.11 \approx 4.25$
Solutions for $f)$ : $x \approx 1.11, 4.25$ .

STEP 5

For $\sin 2x = \frac{1}{\sqrt{2}}$ , recognize that $\frac{1}{\sqrt{2}} = \sin\left(\frac{\pi}{4}\right)$ .
Thus, $2x = \frac{\pi}{4} + 2k\pi$ or $2x = \pi - \frac{\pi}{4} + 2k\pi$ .
Solve for $x$ :
$x = \frac{\pi}{8} + k\pi \quad \text{or} \quad x = \frac{3\pi}{8} + k\pi$
Within $0 \leq x \leq 2\pi$ , the solutions are:
$x \approx 0.39, 1.96, 3.53, 5.11$
Solutions for $a)$ : $x \approx 0.39, 1.96, 3.53, 5.11$ .

STEP 6

For $\sin 3x = -\frac{\sqrt{3}}{2}$ , recognize that $-\frac{\sqrt{3}}{2} = \sin\left(-\frac{\pi}{3}\right)$ .
Thus, $3x = -\frac{\pi}{3} + 2k\pi$ or $3x = \pi + \frac{\pi}{3} + 2k\pi$ .
Solve for $x$ :
$x = -\frac{\pi}{9} + \frac{2k\pi}{3} \quad \text{or} \quad x = \frac{4\pi}{9} + \frac{2k\pi}{3}$
Within $0 \leq x \leq 2\pi$ , the solutions are:
$x \approx 1.05, 2.09, 3.67, 4.71, 6.28$
Solutions for $c)$ : $x \approx 1.05, 2.09, 3.67, 4.71, 6.28$ .

STEP 7

For $\cos 2x = -\frac{1}{2}$ , recognize that $-\frac{1}{2} = \cos\left(\frac{2\pi}{3}\right)$ .
Thus, $2x = \frac{2\pi}{3} + 2k\pi$ or $2x = -\frac{2\pi}{3} + 2k\pi$ .
Solve for $x$ :
$x = \frac{\pi}{3} + k\pi \quad \text{or} \quad x = -\frac{\pi}{3} + k\pi$
Within $0 \leq x \leq 2\pi$ , the solutions are:
$x \approx 1.05, 2.09, 4.19, 5.24$
Solutions for $e)$ : $x \approx 1.05, 2.09, 4.19, 5.24$ .

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