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3 years ago

Solve tan x =cot x for 0 < x < pi

Mathematics

1 answer:

NeX [460]3 years ago

5 0

tan(x) = cot(x)

tan(x) = 1 / tan(x)

tan²(x) = 1

tan²(x) - 1 = 0

(tan(x) - 1) (tan(x) + 1) = 0

tan(x) - 1 = 0 or tan(x) + 1 = 0

tan(x) = 1 or tan(x) = -1

x = π/4 + nπ or x = -π/4 + nπ

where in both cases, n is any integer.

In the interval 0 < x < π, we get solutions:

• x = π/4 from the first family (when n = 0)

• x = 3π/4 from the second family (when n = 1)

So the equation has solutions x = π/4 or x = 3π/4.

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3 years ago

5x3 + 14x2 + 9x help

Cerrena [4.2K]

<h3>Answer: x(x+1)(5x+9) </h3>

===================================================

Work Shown:

5x^3 + 14x^2 + 9x

x( 5x^2 + 14x + 9 )

To factor 5x^2 + 14x + 9, we could use the AC method and guess and check our way to getting the correct result.

A better way in my opinion is to solve 5x^2 + 14x + 9 = 0 through the quadratic formula

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(14)\pm\sqrt{(14)^2-4(5)(9)}}{2(5)}\\\\x = \frac{-14\pm\sqrt{16}}{10}\\\\x = \frac{-14\pm4}{10}\\\\x = \frac{-14+4}{10} \ \text{ or } \ x = \frac{-14-4}{10}\\\\x = \frac{-10}{10} \ \text{ or } \ x = \frac{-18}{10}\\\\x = -1 \ \text{ or } \ x = \frac{-9}{5}\\\\$

Then use those two solutions to find the factorization

x = -1 or x = -9/5

x+1 = 0 or 5x = -9

x+1 = 0 or 5x+9 = 0

(x+1)(5x+9) = 0

So we have shown that 5x^2 + 14x + 9 factors to (x+1)(5x+9)

-----------

Overall,

5x^3 + 14x^2 + 9x

factors to

x(x+1)(5x+9)

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