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Question Number 152094 by mnjuly1970 last updated on 25/Aug/21

Answered by Olaf_Thorendsen last updated on 25/Aug/21

7cosx+4sinx+5 = 0 (1) Let t = tan(x/2) (1) : 7((1−t^2 )/(1+t^2 ))+4((2t)/(1+t^2 ))+5 = 0 7(1−t^2 )+8t+5(1+t^2 ) = 0 t^2 −4t−6 = 0 (t−2)^2 = 10 t = 2±(√(10)) (= tan(α/2) and tan(β/2)) cot(α/2)+cot(β/2) = (1/(2−(√(10))))+(1/(2+(√(10)))) = −(2/3)

$$\mathrm{7cos}{x}+\mathrm{4sin}{x}+\mathrm{5}\:=\:\mathrm{0}\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{t}\:=\:\mathrm{tan}\frac{{x}}{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\::\:\mathrm{7}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{4}\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{5}\:=\:\mathrm{0} \\ $$$$\mathrm{7}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+\mathrm{8}{t}+\mathrm{5}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:=\:\mathrm{0} \\ $$$${t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{6}\:=\:\mathrm{0} \\ $$$$\left({t}−\mathrm{2}\right)^{\mathrm{2}} \:=\:\mathrm{10} \\ $$$${t}\:=\:\mathrm{2}\pm\sqrt{\mathrm{10}}\:\left(=\:\mathrm{tan}\frac{\alpha}{\mathrm{2}}\:\mathrm{and}\:\mathrm{tan}\frac{\beta}{\mathrm{2}}\right) \\ $$$$\mathrm{cot}\frac{\alpha}{\mathrm{2}}+\mathrm{cot}\frac{\beta}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{10}}}+\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{10}}}\:=\:−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 25/Aug/21

vrateful sit olaf..

$${vrateful}\:{sit}\:{olaf}.. \\ $$

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