Question-170084

Question Number 170084 by Beginner last updated on 16/May/22

Commented by cortano1 last updated on 16/May/22

sin x=3((1/2) cos x+(1/2)(√3) sin x) sin x= (3/2) cos x +(3/2)(√3) sin x (((2−3(√3))/2))sin x = (3/2) cos x tan x = (3/(2−3(√3))) = ((3(2+3(√3)))/(4−27)) tan x = −((6+9(√3))/(23))

$$\:\:\:\mathrm{sin}\:{x}=\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{cos}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x}\right) \\ $$$$\:\:\:\mathrm{sin}\:{x}=\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{cos}\:{x}\:+\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x} \\ $$$$\:\left(\frac{\mathrm{2}−\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\mathrm{sin}\:{x}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{cos}\:{x} \\ $$$$\:\:\:\mathrm{tan}\:{x}\:=\:\frac{\mathrm{3}}{\mathrm{2}−\mathrm{3}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{3}\left(\mathrm{2}+\mathrm{3}\sqrt{\mathrm{3}}\right)}{\mathrm{4}−\mathrm{27}} \\ $$$$\:\:\:\mathrm{tan}\:{x}\:=\:−\frac{\mathrm{6}+\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{23}} \\ $$

Commented by mathlove last updated on 16/May/22

$${thanks} \\ $$

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Question-170084

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