+14917 If \(\dfrac{\sin^4 x}2 + \dfrac{\cos^4 x}3 = \dfrac15 \) , find \(tan^2x.\)
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\(\frac{1}{2}(sin^2x)^2+\frac{1}{3}(cos^2x)^2=\frac{1}{5}\\ \frac{1}{2}(sin^2x)^2+\frac{1}{3}(1-sin^2x)^2=\frac{1}{5}\)
\(sin^2x =u\)
\(\frac{1}{2}u^2+\frac{1}{3}(1-u)^2=\frac{1}{5}\\ \frac{1}{2}u^2+\frac{1}{3}(1-2u+u^2)=\frac{1}{5}\\ \frac{1}{2}u^2+\frac{1}{3}-\frac{2}{3}u+\frac{1}{3}u^2=\frac{1}{5}\)
\(\frac{1}{2\cdot 3\cdot 5}\cdot (15u^2+10-20u+10u^2)=\frac{6}{2\cdot 3\cdot 5}\)
\(25u^2-20u+4=0\)
\(u = {20 \pm \sqrt{20^2-4\cdot 25\cdot 4} \over 2\cdot 25}\\ u=\frac{1}{50}\cdot (20\pm 0)\\ \color{blue}u=0.4\)
\(sin^2x=\frac{tan^2x}{1+tan^2x}\)
\(tan^2x=v\)
\(\frac{v}{1+v}=u\\ v=u+uv\\ v-uv=u\\ \color{blue}v=\frac{u}{1-u} \)
\(tan^2x=\frac{0.4}{1-0.4}\)
\(tan^2x=\frac{2}{3}\)
Thanks heureka!
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