# Answer to Question #1755 in Trigonometry for Isha

Question #1755

Solve this trigonometric equation.give all positive values of the angle between 0deg and 360deg that will satisfy.<br>Give any approximate value to the nearest minute only.

<br>& tan (x+15) = 3 tanx

<br>& tan (x+15) = 3 tanx

Expert's answer

tan(x+15) = (tanx + tan15)/(1-tanx tan15)

(tanx + tan15)/(1-tanx tan15) = 3 tanx

tanx + tan 15 = 3 tanx - 3tan2x tan15

3 tan

tan15 = 0.27, denote tanx as t

0.81 t

D = 0.1252

t

t

tan (x

x

tan(x

x

Answer: 59º 5', 239º 5', 38º 40', 218º 40'

(tanx + tan15)/(1-tanx tan15) = 3 tanx

tanx + tan 15 = 3 tanx - 3tan2x tan15

3 tan

^{2}x tan15 - 2tanx + tan15 = 0tan15 = 0.27, denote tanx as t

0.81 t

^{2 }- 2t + 0.27 = 0D = 0.1252

t

_{1}= 1.67t

_{2 }= 0.80tan (x

_{1}) = 1.67x

_{1}= arctan (1.67) + πn = 59.087 + 180n = 59º 5'+ 180n, n=0,1,2,..tan(x

_{2}) = 0.80x

_{2}= arctan(0.80) + πn = 38.66 + 180n = 38º 40' + 180n, n=0,1,2,..Answer: 59º 5', 239º 5', 38º 40', 218º 40'

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