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

which term describes the soviet policy the expanded freedom of the press and allowed criticism of the communist party during the cold warA: containment B:Perest
Find the real square root of 0.81
What is fraction that is equivalent to 3/8
To save time, you can approximate the initial mass of the solid to the nearest ±1 g. For example, if you are asked to add 14.3 g of copper, add between 13 g and
An arithmetic series a consists of consecutive integers that are multiples of 4 what is the sum of the first 9 terms of this sequence if the first term is 0
Find the condition that the zeros of the polynomial f(x) = x^3+3px^2+3px+r may be in A.P.
How many molecules are present in 3 mols of silicon dioxide (sio2)?
A circle has a circumference of 7.850 units. What is its radius?
If the speed of a particle triples ,by what factor does its kinetic energy increase?
suggest why placing wheels under a heavy box reduces the neccessary force required to push it along at a constant speed.