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Question Number 165637 by daus last updated on 05/Feb/22

37tanx =11tan3x  solve it

$$\mathrm{37}{tanx}\:=\mathrm{11}{tan}\mathrm{3}{x} \\ $$$${solve}\:{it} \\ $$

Answered by mr W last updated on 06/Feb/22

37 tan x=11×((3 tan x−tan^3  x)/(1−3 tan^2  x))  ⇒tan x=0 ⇒x=kx  ⇒37=11×((3−tan^2  x)/(1−3 tan^2  x))  37−3×37 tan^2  x=3×11−11 tan^2  x  25 tan^2  x=1  ⇒tan x=±(1/5) ⇒x=kπ±tan^(−1) (1/5)

$$\mathrm{37}\:\mathrm{tan}\:{x}=\mathrm{11}×\frac{\mathrm{3}\:\mathrm{tan}\:{x}−\mathrm{tan}^{\mathrm{3}} \:{x}}{\mathrm{1}−\mathrm{3}\:\mathrm{tan}^{\mathrm{2}} \:{x}} \\ $$$$\Rightarrow\mathrm{tan}\:{x}=\mathrm{0}\:\Rightarrow{x}={kx} \\ $$$$\Rightarrow\mathrm{37}=\mathrm{11}×\frac{\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \:{x}}{\mathrm{1}−\mathrm{3}\:\mathrm{tan}^{\mathrm{2}} \:{x}} \\ $$$$\mathrm{37}−\mathrm{3}×\mathrm{37}\:\mathrm{tan}^{\mathrm{2}} \:{x}=\mathrm{3}×\mathrm{11}−\mathrm{11}\:\mathrm{tan}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{25}\:\mathrm{tan}^{\mathrm{2}} \:{x}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{tan}\:{x}=\pm\frac{\mathrm{1}}{\mathrm{5}}\:\Rightarrow{x}={k}\pi\pm\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}} \\ $$

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