Laipsninė lygtis su sinusu ir kosinusu

[tex]5^{cos^{2}x}+5^{sin^{2}x}=6[/tex]

[tex]5^{1-\sin^2x}+5^{\sin^2x}=6\implies \dfrac{5}{5^{\sin^2x}}+5^{\sin^2x}=6[/tex]
[tex]t=5^{\sin^2x}[/tex]
[tex]\dfrac{5}{t}+t=6|\cdot t≠0\implies t^2-6t+5=0\implies t=1,\space t=5.[/tex]
[tex]5^{\sin^2x}=1\implies 5^{\sin^2x}=5^0\implies \sin^2x=0\implies \sin x=0\implies x=\pi k,k∈\mathbb{Z}[/tex]
[tex]5^{\sin^2x}=5\implies 5^{\sin^2x}=5^1\implies \sin^2x=1\implies \sin x=1,\space \sin x=-1\implies \\x=\dfrac{\pi}{2}+ \pi k,k∈\mathbb{Z}[/tex]
Atsakymas: [tex]\dfrac{\pi k}{2},k∈\mathbb{Z}[/tex]

sin²x geriau pertvarkyti į (1-cos(2x))/2. Tada išvengiama +,-1. x=π/2+πn, n∈Z. Atsakymą gaunu tokį pat.