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# trig

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If $$\dfrac{\sin^4 x}2 + \dfrac{\cos^4 x}3 = \dfrac15$$, find $$\tan^2 x$$

Jul 7, 2020

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If $$\dfrac{\sin^4 x}2 + \dfrac{\cos^4 x}3 = \dfrac15$$ ,  find  $$tan^2x.$$

Hello Guest!

$$\frac{1}{2}(sin^2x)^2+\frac{1}{3}(cos^2x)^2=\frac{1}{5}\\ \frac{1}{2}(sin^2x)^2+\frac{1}{3}(1-sin^2x)^2=\frac{1}{5}$$

$$sin^2x =u$$

$$\frac{1}{2}u^2+\frac{1}{3}(1-u)^2=\frac{1}{5}\\ \frac{1}{2}u^2+\frac{1}{3}(1-2u+u^2)=\frac{1}{5}\\ \frac{1}{2}u^2+\frac{1}{3}-\frac{2}{3}u+\frac{1}{3}u^2=\frac{1}{5}$$

$$\frac{1}{2\cdot 3\cdot 5}\cdot (15u^2+10-20u+10u^2)=\frac{6}{2\cdot 3\cdot 5}$$

$$25u^2-20u+4=0$$

$$u = {20 \pm \sqrt{20^2-4\cdot 25\cdot 4} \over 2\cdot 25}\\ u=\frac{1}{50}\cdot (20\pm 0)\\ \color{blue}u=0.4$$

$$sin^2x=\frac{tan^2x}{1+tan^2x}$$

$$tan^2x=v$$

$$\frac{v}{1+v}=u\\ v=u+uv\\ v-uv=u\\ \color{blue}v=\frac{u}{1-u}$$

$$tan^2x=\frac{0.4}{1-0.4}$$

$$tan^2x=\frac{2}{3}$$

Thanks heureka!

!

Jul 7, 2020
edited by asinus  Jul 7, 2020
edited by asinus  Jul 7, 2020
edited by asinus  Jul 7, 2020
edited by asinus  Jul 7, 2020