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04-02-2021
Mathematics

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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=

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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:

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