MOHON BANTUANNYA............... himpunan penyelesaian dari tan²x - tanx + 1 = 0 adalah.... Question from @HengkyW

October 2018 1 69 Report

MOHON BANTUANNYA...............
himpunan penyelesaian dari tan²x - $\frac{4}{3} \sqrt{3}$ tanx + 1 = 0 adalah...

izharulhaq78 $\displaystyle \text{misal :}\\\tan x=a\\\\\tan^2x-\frac{4}{3}\sqrt3\tan x+1=0\\a^2-\frac{4}{3}\sqrt3a+1=0\\\\a_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\a_{1,2}=\frac{-(-\frac{4}{3}\sqrt3)\pm\sqrt{(-\frac{4}{3}\sqrt3)^2-4(1)(1)}}{2(1)}\\a_{1,2}=\frac{\frac{4}{3}\sqrt3\pm\sqrt{(\frac{16}{9}\times3)-4}}{2}$
$\displaystyle a_{1,2}=\frac{\frac{4}{3}\sqrt3\pm\sqrt{\frac{16}{3}-\frac{12}{3}}}{2}\\a_{1,2}=\frac{\frac{4}{3}\sqrt3\pm\sqrt{\frac{4}{3}}}{2}\\a_{1,2}=\frac{\frac{4}{3}\sqrt3\pm\frac{2}{\sqrt3}}{2}\\a_{1,2}=\frac{\frac{4}{3}\sqrt3\pm\frac{2}{3}\sqrt3}{2}\\a_{1}=\frac{\frac{4}{3}\sqrt3+\frac{2}{3}\sqrt3}{2}\text{ atau }a_{2}=\frac{\frac{4}{3}\sqrt3-\frac{2}{3}\sqrt3}{2}\\a_{1}=\frac{\frac{6}{3}\sqrt3}{2}\text{ atau }a_{2}=\frac{\frac{2}{3}\sqrt3}{2}$
$\displaystyle a_{1}=\frac{2\sqrt3}{2}\text{ atau }a_{2}=\frac{1}{3}\sqrt3\\a_{1}=\sqrt3\text{ atau }a_{2}=\frac{1}{\sqrt3}\\\\a_1=\sqrt3\\\tan x=\sqrt3\\\arctan\sqrt3=x\\60^o=x\\\\\text{atau}\\\\a_{2}=\frac{1}{\sqrt3}\\\tan x=\frac{1}{\sqrt3}\\\arctan\frac{1}{\sqrt3}=x\\30^o=x\\\\\boxed{HP=\{30^o,60^o\}}$

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Jika sinβ + cosβ = , tentukan sinβ.cosβ

Tentukan nilai dari 3tan²x-sec²x-5=0

Nilai dari sin() adlah....?

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