[tex]1+tg^2x=8sin^2x \\ 1+ \frac{sin^2x}{cos^2x}=8sin^2x \;\;\;\;|:sin^2x \\ \\ \frac{1}{sin^2x} + \frac{ \frac{sin^2x}{cos^2x}}{sin^2x} =8 \\ \\ \frac{1}{sin^2x} + \frac{sin^2x}{cos^2x}* \frac{1}{sin^2x} =8 \\ \\ \frac{1}{sin^2x} + \frac{1}{cos^2x} =8 \;\;\;\;\;\;\; \text{ aducem la acelasi numitor} \\ \\ \frac{cos^2x}{sin^2x*cos^2x} + \frac{sin^2x}{sin^2x*cos^2x} =8 \\ \\ \frac{sin^2x+cos^2x}{sin^2x*cos^2x}=8 \;\;\;\;\;dar \;\;\;sin^2x+cos^2x = 1[/tex]
[tex]\frac{1}{sin^2x*cos^2x}=8 \\ \\ sin^2x*cos^2x = \frac{1}{8} \\ \\ (sin\,x\;cos\,x)^2 = \frac{1}{8} \\ . \;\;\aplicam \;formula: \; sin\,2\alpha=2sin\,\alpha \;cos\,\alpha\;\; =\ \textgreater \ \;\; sin\,\alpha \;cos\,\alpha = \frac{sin\,2\alpha}{2} \\ \\ (\frac{sin\,2x}{2} )^{2}=\frac{1}{8} \\ \\ sin^22x=\frac{4}{8} \\ \\ sin^22x=\frac{1}{2} \\ \\ sin\,2x = \sqrt{ \frac{1}{2} } \\ sin\,2x = { \frac{ \sqrt2}{2} } \\ \\ 2x =arcsin\frac{ \sqrt2}{2} } \\ \\ \text{Sinusul este pozitiv in cadranul 1 si 2} => avem \;2 \;solutii: \\ \\ 2x_1 = 45^o = \frac{\pi}{4} \;\;(\text{Deoarece avem interval impus renuntam la }"+2k\pi") \\ \\ 2x_2 = 135^o = \frac{3\pi}{4} [/tex]
[tex]x_1 = 22,5^o = \frac{\pi}{8} \;\; \text{Solutie eliminata deoarece nu apartine intervalului dat} \\ \\ x_2 =\boxed{ 67,5^o = \frac{3\pi}{8}} [/tex]
