SOLUTION: Use inverse functions where needed to find all solutions of the equation in the interval [0, 2π)
tan^2 x − 5 tan x − 6 = 0
sec^2 x − 4 tan x = −2
2
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-> SOLUTION: Use inverse functions where needed to find all solutions of the equation in the interval [0, 2π)
tan^2 x − 5 tan x − 6 = 0
sec^2 x − 4 tan x = −2
2
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Question 847345: Use inverse functions where needed to find all solutions of the equation in the interval [0, 2π)
tan^2 x − 5 tan x − 6 = 0
sec^2 x − 4 tan x = −2
2 cos^2 x + 7 sin x = 5
Cot^2 x − 25 = 0
tan^2 x + tan x − 6 = 0
2 cos^2 x + 7 sin x = 5
You can put this solution on YOUR website! Use inverse functions where needed to find all solutions of the equation in the interval [0, 2π)
tan^2 x − 5 tan x − 6 = 0
Factor:
(tan-6)(t+1) = 0
tan(x) = 6 or tan(x) = -1
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x = arctan(6) or tan(x) = -1
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x = 1.4056.. or x = (pi+1.4056) or x = (3/4)pi or (7/4)pi
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sec^2 x − 4 tan x = −2
(1+tan^2(x))-4tan(x) = -2
tan^2(x) - 4tan(x) +3 = 0
(tan(x)-3)(tan(x)-1) = 0
tan(x) = 3 or tan(x) = 1
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x = arctan(3) or x = arctan(1)
x = 1.23 or x = pi+1.23 or x = (pi/4) or x = (5/4)pi
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Cheers,
Stan H.
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