Посмотри решение, может где найдешь ошибку....
нашла одну ошибку, исправила, но все равно не решается как надо...
(если узнаешь правильное решение сообщи, если не трудно)
вроде так, не уверен со знаком после 1й скобки
c(c+6)(3+d)(3-d)
Cosx -√3 Sinx = 2
Cos²x/2 -Sin²x/2 -√3*2Sinx/2Cosx/2 = 2*(Sin²x/2 + Cos²x/2)
Cos²x/2 -Sin²x/2 -√3*2Sinx/2Cosx/2 = 2Sin²x/2 + 2Cos²x/2
Cos²x/2 -Sin²x/2 -√3*2Sinx/2Cosx/2 -Sin²x/2 -2 Cos²x/2=0
-2Sin²x/2 -2√3SinxCosx -Cos²x/2 = 0 | : (-Cos²x/2)
tg²x/2 +2√3 tgx/2 +1 = 0
tgx/2 = t
t² + 2√3 t +1 = 0
t = -√3 +-√(3-1) = -√3 +-√2
tgx/2 = -√3 +-√2
x/2 = arctg(-√3 +-√2) + πk , k ∈Z
x = 2arctg(-√3 +-√2) + 2πk , k ∈Z
tg (22°30') > 0 ; сtg (22°30') > 0 - первая четверть
![tg (22\textdegree 30')=\sqrt{\dfrac{1-\cos (45\textdegree)}{1+\cos (45\textdegree)}}=\sqrt\dfrac{1-\frac{\sqrt2}2}{1+\frac{\sqrt2}2}}=\\\\\\=\sqrt{\dfrac{2-\sqrt2}{2+\sqrt2}}=\sqrt{\dfrac{(2-\sqrt2)^2}{(2+\sqrt2)\cdot (2-\sqrt2)}}=\\\\\\=\sqrt{\dfrac{4-4\sqrt2+2}{2}}\boldsymbol{=\sqrt{3-2\sqrt2}}](https://tex.z-dn.net/?f=tg+%2822%5Ctextdegree+30%27%29%3D%5Csqrt%7B%5Cdfrac%7B1-%5Ccos+%2845%5Ctextdegree%29%7D%7B1%2B%5Ccos+%2845%5Ctextdegree%29%7D%7D%3D%5Csqrt%5Cdfrac%7B1-%5Cfrac%7B%5Csqrt2%7D2%7D%7B1%2B%5Cfrac%7B%5Csqrt2%7D2%7D%7D%3D%5C%5C%5C%5C%5C%5C%3D%5Csqrt%7B%5Cdfrac%7B2-%5Csqrt2%7D%7B2%2B%5Csqrt2%7D%7D%3D%5Csqrt%7B%5Cdfrac%7B%282-%5Csqrt2%29%5E2%7D%7B%282%2B%5Csqrt2%29%5Ccdot+%282-%5Csqrt2%29%7D%7D%3D%5C%5C%5C%5C%5C%5C%3D%5Csqrt%7B%5Cdfrac%7B4-4%5Csqrt2%2B2%7D%7B2%7D%7D%5Cboldsymbol%7B%3D%5Csqrt%7B3-2%5Csqrt2%7D%7D)
![ctg (22\textdegree 30')=\sqrt{\dfrac{1+\cos (45\textdegree)}{1-\cos (45\textdegree)}}=\sqrt\dfrac{1+\frac{\sqrt2}2}{1-\frac{\sqrt2}2}}=\\\\\\=\sqrt{\dfrac{2+\sqrt2}{2-\sqrt2}}=\sqrt{\dfrac{(2+\sqrt2)^2}{(2-\sqrt2)\cdot (2+\sqrt2)}}=\\\\\\=\sqrt{\dfrac{4+4\sqrt2+2}{2}}\boldsymbol{=\sqrt{3+2\sqrt2}}](https://tex.z-dn.net/?f=ctg+%2822%5Ctextdegree+30%27%29%3D%5Csqrt%7B%5Cdfrac%7B1%2B%5Ccos+%2845%5Ctextdegree%29%7D%7B1-%5Ccos+%2845%5Ctextdegree%29%7D%7D%3D%5Csqrt%5Cdfrac%7B1%2B%5Cfrac%7B%5Csqrt2%7D2%7D%7B1-%5Cfrac%7B%5Csqrt2%7D2%7D%7D%3D%5C%5C%5C%5C%5C%5C%3D%5Csqrt%7B%5Cdfrac%7B2%2B%5Csqrt2%7D%7B2-%5Csqrt2%7D%7D%3D%5Csqrt%7B%5Cdfrac%7B%282%2B%5Csqrt2%29%5E2%7D%7B%282-%5Csqrt2%29%5Ccdot+%282%2B%5Csqrt2%29%7D%7D%3D%5C%5C%5C%5C%5C%5C%3D%5Csqrt%7B%5Cdfrac%7B4%2B4%5Csqrt2%2B2%7D%7B2%7D%7D%5Cboldsymbol%7B%3D%5Csqrt%7B3%2B2%5Csqrt2%7D%7D)
Проверка :
![tg (22\textdegree 30')\cdot ctg (22\textdegree 30')=\\\\=\sqrt{3-2\sqrt2}\cdot \sqrt{3+2\sqrt2}=3^2-(2\sqrt 2)^2=1](https://tex.z-dn.net/?f=tg+%2822%5Ctextdegree+30%27%29%5Ccdot+ctg+%2822%5Ctextdegree+30%27%29%3D%5C%5C%5C%5C%3D%5Csqrt%7B3-2%5Csqrt2%7D%5Ccdot+%5Csqrt%7B3%2B2%5Csqrt2%7D%3D3%5E2-%282%5Csqrt+2%29%5E2%3D1)
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Использованы формулы
![tg^2\dfrac {\alpha }2=\dfrac{1-\cos \alpha }{1+ \cos \alpha }\\\\ctg^2\dfrac {\alpha }2=\dfrac{1+\cos \alpha }{1- \cos \alpha }\\\\a^2-b^2=(a-b)(a+b)](https://tex.z-dn.net/?f=tg%5E2%5Cdfrac+%7B%5Calpha+%7D2%3D%5Cdfrac%7B1-%5Ccos+%5Calpha+%7D%7B1%2B+%5Ccos+%5Calpha+%7D%5C%5C%5C%5Cctg%5E2%5Cdfrac+%7B%5Calpha+%7D2%3D%5Cdfrac%7B1%2B%5Ccos+%5Calpha+%7D%7B1-+%5Ccos+%5Calpha+%7D%5C%5C%5C%5Ca%5E2-b%5E2%3D%28a-b%29%28a%2Bb%29)
1) =х√36х^6= x*I6x³I =x*-6x=-6x². где х<0
2) =2m³ √(9n²/m²)= 2m²* I 3n/m I= 2m²*(-2n/m)=-4mn. где n<0
I I это модуль