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3 years ago

5. Find a counterexample to show that the following con-

Mathematics

1 answer:

Bess [88]3 years ago

7 0

Answer: see below

<u>Step-by-step explanation:</u>

Any value x such that |x| < 1 will make the conjecture false

<em>In simpler words, let x be a fraction between 0 and 1.</em>

One example: Let x = $\dfrac{1}{2}$

Then x⁴ = $\bigg(\dfrac{1}{2}\bigg)^4$

= $\dfrac{1}{16}$

$\dfrac{1}{16}$ < $\dfrac{1}{2}$

so the conjecture is false.

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How do I solve: 2 sin (2x) - 2 sin x + 2√3 cos x - √3 = 0

ziro4ka [17]

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By the double angle identity of sines:

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Rewrite the original equation with this identity:

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

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

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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:

$\displaystyle \sin x = -\frac{\sqrt{3}}{2}$ , which corresponds to $\displaystyle x = -\frac{\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = -\frac{2\, \pi}{3} + 2\, k\, \pi$ .
$\displaystyle \cos x = \frac{1}{2}$ , which corresponds to $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ and $\displaystyle x = -\frac{\pi}{3} + 2\, k \, \pi$ .

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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$\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ .

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3 years ago