Find two solutions of each equation? Give your anwser in radians (0 < \theta < 2 π). Dont use no caculator.

Find two solutions of each equation? Give your anwser in radians (0 < $\theta$ < 2 π). Dont use no caculator.

tan $\theta$ = 1

cot $\theta$ = $- \sqrt{}3$

asked Sep 5, 2018 in TRIGONOMETRY by rockstar Apprentice
reshown Sep 5, 2018 by bradely

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1 Answer

tan $\theta$ = 1

tan $\theta$ = tan(π/4)

$\theta$ = nπ+π/4

At n = 1

$\theta$ =π+π/4 =5π/4

cot $\theta$ =-√3

tan $\theta$ =-1/√3

tan $\theta$ =tan(-π/6)

$\theta$ = nπ-π/6

At n = 1

$\theta$ = π-π/6=5π/6

At n = 2

$\theta$ = 2π-π/6=11π/6

$\theta$ =5π/6 and 11π/6

answered Sep 14, 2018 by lilly Expert
reshown Sep 14, 2018 by bradely

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