trigonometric equation on the intervals [0,2pi).

Question

use trigonometric, identities, algebraic, message, and inverse trigonometric functions, as necessary, just saw the following trigonometric equation on the intervals [0,2pi). round your answer to four decimal places, if necessary. If there is no solution, indicate no solution. 5cot(x)+5=0

Answer 1

5cot(x) + 5 = 0

Divide both sides by 5:

cot(x) + 1 = 0

Subtract 1 from both sides:

cot(x) = -1

Take the cot^-1 of both sides:

x = 3π/4, 7π/4

x ≈ 2.3562, 5.4978

Answer 2

Hi Alyson!

This one's pretty straightforward.

Divide through by 5:

cot(x) + 1 = 0

cos(x) / sin(x) = -1

cos(x) = - sin(x)

which is true for 3 pi/4 + any multiple of pi

Answer 3

5 cot(x) + 5 = 0

5cot(x) = -5 Move constant to right side of =.

cot (x) = -1 Divide both sides by 5

1 = -1 identity. cot X = 1

tan (x) tan(x)

-tan(x) = 1

tan (x) = -1

x = 3π/4, 7π/4 Unit Circle