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3 years ago

Find a numerical value if one trigonometric function of x if tan2x-sin2x/sin2x=5

Mathematics

1 answer:

seropon [69]3 years ago

4 0

Answer:

The values of x that satisfy the given equation are:

x1 = 1.183 + nπ

x2 = -1.183 + nπ

Step-by-step explanation:

Given tan²x - sin²x/sin²x = 5

Simplifying this, we have

tan²x - 1 = 5

Adding 1 to both sides, we have

tan²x = 6

Because tan²x = (tanx)², we can write as

(tanx)² = 6

Taking square roots of both sides, we have.

tanx = ±√6

x = arctan(±√6) + nπ

≈ 1.183 + nπ or -1.183 + nπ

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Aidan is participating in an 8 mile walk for his favorite charity. The organizers used a coordinate grid to plot the course. The

Travka [436]

Answer:

Aidan is 2 miles far from the ending point when he reaches the water station.

Step-by-step explanation:

The locations of the starting point, water station and ending point are (3, 1), (3, 7) and (3, 9), all expressed in miles. First we determine the distances between starting and ending points and between starting point and water station by the Pythagorean Theorem:

From starting point to ending point:

$D = \sqrt{(3-3)^{2}+(9-1)^{2}}$ (Eq. 1)

$D = 8\,mi$

From starting point to water station:

$d = \sqrt{(3-3)^{2}+(7-1)^{2}}$ (Eq. 2)

$d = 6\,mi$

The distance between the water station and the ending point is:

$s = D-d$ (Eq. 3)

$s = 8\,mi-6\,mi$

$s = 2\,mi$

Hence, Aidan is 2 miles far from the ending point when he reaches the water station.

3 0

3 years ago

The square root of 5 is between which two integers

Verdich [7]

$2$

6 0

3 years ago

The slope-intercept form of the equation of a line that passes through point (-3, 8) is y = -2/3x + 6. What is the point-

Elena-2011 [213]

$y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{3}}x+6\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so hmmm then we know the slope of that line is -2/3, so we're really looking for the point-slope form of a line with a slope of -2/3 and that passes through (-3 , 8)

$(\stackrel{x_1}{-3}~,~\stackrel{y_1}{8})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{8}=\stackrel{m}{-\cfrac{2}{3}}(x-\stackrel{x_1}{(-3)})\implies y-8=-\cfrac{2}{3}(x+3)$