Тригонометрични неравенства - 11ти клас, екстра гадни :>

[Phenomenon]

На пръв поглед са елементарни, само че идея си нямам..
1) sin³xsin(π/2-3x) + cos³xcos(π/2-3x)>3√3/8
2) 2sin²x - sinx + sin3x<1
3)cotgx + cotg(x+π/2) + 2cotg(x+π/2)>0

Благодаря предварително на помогналите

[Реклама]

[_sssss]

[tex]2. \; 2sin^2 x - sinx + 3sinx - 4sin^3 x -1 <0 \\ sinx=u \\ 4u^3 - 2u^2 - 2u + 1 > 0 \\ (2u-1)(2u^2-1)>0 \\ u\in(-\frac{\sqrt{2}}{2}; \frac{1}{2})\cup(\frac{\sqrt{2}}{2}; +\infty) \\ x\in(-\frac{\pi}{4}+2k\pi; \frac{\pi}{6}+2k\pi)\cup(\frac{\pi}{4}+2k\pi; \frac{3\pi}{4}+2k\pi)\cup(\frac{5\pi}{6}+2k\pi; \frac{5\pi}{4}+2k\pi);[/tex]
?

[mkmarinov]

[tex]sin^3x.cos3x+cos^3x.sin3x=sin^3x(4cos^3x-3cosx)+cos^3x(3sinx-4sin^3x)=\\ 4sin^3xcos^3x-3sin^3xcosx+3cos^3xsinx-4sin^3xcos^3x=3sinxcosx(cos^2x-sin^2x)=\frac{3}{2}sin2xcos2x=\frac{3}{4}sin4x[/tex]
[tex]\frac{3}{4}sin4x>\frac{3\sqrt{3}}{8}[/tex]
[tex]sin4x>\frac{\sqrt{3}}{2}[/tex]

3)
[tex]cotgx+3cotg(x+\pi /2)=cotgx+3tgx=\frac{1}{tgx}+3tgx=\frac{3tg^2x+1}{tgx}[/tex]
[tex]\frac{3tg^2x+1}{tgx}>0[/tex]
[tex]tgx>0[/tex]