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Neko [114]
4 years ago
6

Find an equation of the tangent line to the graph of the function f(x) = 2sinx at the point where x = pi/3.

Mathematics
1 answer:
BigorU [14]4 years ago
3 0
The equation of the tangent line of f(x) is given by y = f'(x)x + c
f(x) = 2sin x
f'(x) = 2cos x

At x = pi/3
f'(pi/3) = 2 cos(pi/3) = 1
f(pi/3) = 2 sin(pi/3) = √3
√3 = 1(π/3) + c
c = √3 - π/3

Therefore, required equation is
y = x + √3 - π/3
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How do I solve: 2 sin (2x) - 2 sin x + 2√3 cos x - √3 = 0
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Answer:

\displaystyle x = \frac{\pi}{3} +k\, \pi or \displaystyle x =- \frac{\pi}{3} +2\,k\, \pi, where k is an integer.

There are three such angles between 0 and 2\pi: \displaystyle \frac{\pi}{3}, \displaystyle \frac{2\, \pi}{3}, and \displaystyle \frac{4\,\pi}{3}.

Step-by-step explanation:

By the double angle identity of sines:

\sin(2\, x) = 2\, \sin x \cdot \cos x.

Rewrite the original equation with this identity:

2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0.

Note, that 2\, (2\, \sin x \cdot \cos x) and (-2\, \sin x) share the common factor (2\, \sin x). On the other hand, 2\sqrt{3}\, \cos x and (-\sqrt{3}) share the common factor \sqrt[3}. Combine these terms pairwise using the two common factors:

(2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0.

Note the new common factor (2\, \cos x - 1). Therefore:

\left(2\, \sin x + \sqrt{3}\right) \cdot (2\, \cos x - 1) = 0.

This equation holds as long as either \left(2\, \sin x + \sqrt{3}\right) or (2\, \cos x - 1) is zero. Let k be an integer. Accordingly:

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Any x that fits into at least one of these patterns will satisfy the equation. These pattern can be further combined:

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3 years ago
. Use the quadratic formula to solve each quadratic real equation. Round
Liono4ka [1.6K]

Answer:

A. No real solution

B. 5 and -1.5

C. 5.5

Step-by-step explanation:

The quadratic formula is:

\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}, with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}

\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5

So B has two solutions of 5 and -1.5.

Now to C!

\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}

\frac{44}{8} = 5.5

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

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Step-by-step explanation:

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