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Solve the following for 0 Β° ≀ πœƒ ≀ 360Β°

(a) cos π‘₯ + 0.582 = 0

(b) cot π‘₯ = βˆ’0.836

(c) 2𝑠𝑖𝑛2π‘₯ βˆ’ cos π‘₯ = 1

Guest Mar 27, 2017
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(a)

cos x + 0.582 = 0

cos x = 0 - 0.582

x = acos(-0.582)

x β‰ˆ 125.591ΒΊ       and      x β‰ˆ 360ΒΊ - 125.591ΒΊ β‰ˆ 234.409ΒΊ

 

(b)

cot x = -0.836

tan x = -1/0.836

tan x β‰ˆ -1.196

x β‰ˆ atan(-1.196)

x β‰ˆ -50.1ΒΊ

x β‰ˆ -50.1ΒΊ + 360ΒΊ β‰ˆ 309.9ΒΊ       and       x β‰ˆ 309.9ΒΊ - 180ΒΊ β‰ˆ 129.9ΒΊ

hectictar  Mar 27, 2017

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