Question

KUDA
(Kuis mudah)

Diketahui
[tex]\huge \frac{{sin}^{4}x } {2} + \frac{ {cos}^{4}x}{3} = \frac{1}{5} [/tex]
Maka nilai dari tan²x ..........

a) 2
b) 4/5
c) 2/3
d) 3/2​

Answer 1

Verified answer

Jawab:
c) 2/3

Penjelasan:
[tex]\displaystyle\rm\frac{sin^4x}{2}+\frac{cos^4x}{3}=\frac{1}{5}\\\\\frac{(sin^2x)^2}{2}+\frac{(cos^2x)^2}{3}=\frac{1}{5}\\\\\frac{(sin^2x)^2}{2}+\frac{(1-sin^2x)^2}{3}=\frac{1}{5}\\\\sin^2x=u,maka\\\\\frac{u^2}{2}+\frac{(1-u)^2}{3}=\frac{1}{5}\\\\\frac{u^2}{2}+\frac{1-2u+u^2}{3}=\frac{1}{5}\\\\\frac{15u^2}{\not2(\not15)}+\frac{10(1-2u+u^2)}{\not3(\not10)}=\frac{6}{\not5(\not6)}[/tex]
15u² + 10 - 20u + 10u² = 6
25u² - 20u + 4 = 0
(5u)² - 2(5u)(2) + 2² = 0
(5u - 2)² = 0
Akar kembar..
5u - 2 = 0
[tex]\displaystyle u=\frac{2}{5}\\\\\rm sin^2(x)=\frac{2}{5}\\\\cos^2(x)=1-sin^2(x) =1-\frac{2}{5}=\frac{3}{5}\\\\\therefore tan^2(x)=\frac{sin^2(x)}{cos^2(x)}=\frac{2}{5}\div\frac{3}{5}=\frac{2}{\not5}\times\frac{\not5}{3}[/tex]

c) 2/3
(xcvi)

Answer 2

Penjelasan dengan langkah-langkah:

terlampir ya

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