abigailbarnes8242 abigailbarnes8242

04-02-2021
Mathematics

contestada

Solve the equation for x:
sin^2 x – 2 cos x - 2 = 0
for the restriction 0 < x < 2pi

Solve the equation for x sin2 x 2 cos x 2 0 for the restriction 0 lt x lt 2pi class=

Respuesta :

Аноним Аноним

04-02-2021

Answer: x = pi

sin²x - 2cos x - 2 = 0

⇔ 1 - cos²x - 2cos x - 2 = 0

⇔ cos²x + 2cos x + 1 = 0

⇔ (cos x + 1)² = 0

⇒ cos x + 1 = 0

⇔ cos x = -1

⇒ x = pi + k2pi

because 0 ≤ x < 2pi

=> x = pi

Step-by-step explanation:

Answer Link

Otras preguntas

what is a transfer of food goods and disease between the old new worlds columbian

Simplify 17d-5e+13d-7e

how did world war 2 end in a prisoner's story

There are 34 adults and children13 on tour how much money will the group save by visiting the main deck of the observation tower instead of the top deck explain

A man in a photograph is 1.5 inches in height. If the man is 6 ft. tall, what is the scale.

PLEASE HELP What is the mode of the following set of numbers? 51, 73, 82, 51, 73, 103, 73, 125, 82, 125 51 73 125 82

What is decrease £20 by 15%

What is the capital of Maine

Which group convinced President James Madison to go to war with England in 1812? A. land-hungry politicians from the West and South B. angry veterans of the Rev

solve the following quadratic equation (x+3)2+25=0