What is the general solution of the equation sin^2x +2cos^2x +3sinx cosx=0?

     
\frac{-2\left(\sin(x)\right)^{4}-2\left(\cos(x)\right)^{4}+7\cos(x)\left(\sin(x)\right)^{2}+3\left(\cos(x)\right)^{3}-\left(\sin(2x)\right)^{2}+4\cos(2x)-4\cos(x)-3}{\left(\sin(x)\left(-2\cos(x)+3\right)\right)^{2}}
*

How do you simplify \displaystyle\frac{{{\cos{{4}}}{x}-{\cos{{2}}}{x}}}{{{\sin{{4}}}{x}+{\sin{{2}}}{x}}} ?
\displaystyle{\left(\Rightarrow-{\tan{{x}}}\right.} Explanation: \displaystyle\frac{{{\cos{{4}}}{x}-{\cos{{2}}}{x}}}{{{\sin{{4}}}{x}+{\sin{{2}}}{x}}}\displaystyle{\cos{{C}}}-{\cos{{D}}}=-{2}\frac{{\sin{{\left({C}+{D}\right)}}}}{{2}}\frac{{\sin{{\left({C}-{D}\right)}}}}{{2}} ...

Bạn đang xem: What is the general solution of the equation sin^2x +2cos^2x +3sinx cosx=0?


If \displaystyle{\tan{{x}}}=-\frac{{12}}{{5}} , find \displaystyle\frac{{{\left({\cos{{x}}}-{1}\right)}{\left({\cos{{x}}}+{1}\right)}}}{{{\left({\sin{{x}}}-{1}\right)}{\left({\sin{{x}}}+{1}\right)}}} ...
\displaystyle\frac{{{\left({\cos{{x}}}-{1}\right)}{\left({\cos{{x}}}+{1}\right)}}}{{{\left({\sin{{x}}}-{1}\right)}{\left({\sin{{x}}}+{1}\right)}}}=\frac{{144}}{{25}} Explanation:As\displaystyle{\tan{{x}}}=-\frac{{12}}{{5}} ...
How do you prove \displaystyle\frac{{{1}-{\cos{{x}}}}}{{\sin{{x}}}}=\frac{{\sin{{x}}}}{{{1}+{\cos{{x}}}}} ?
Please see below.Explanation: \displaystyle\frac{{{1}-{\cos{{x}}}}}{{\sin{{x}}}} =\displaystyle\frac{{{1}-{\cos{{x}}}}}{{\sin{{x}}}}\times\frac{{{1}+{\cos{{x}}}}}{{{1}+{\cos{{x}}}}} =\displaystyle\frac{{{1}-{{\cos}^{{2}}{x}}}}{{{\sin{{x}}}{\left({1}+{\cos{{x}}}\right)}}} ...

Xem thêm: Các Góc Phần Tư Thứ Nhất Là Bao Nhiêu Độ, Góc Phần Tư Thứ 2


How do you find \displaystyle\frac{{{\cos{{5}}}{x}+{\cos{{4}}}{x}}}{{{\sin{{3}}}{x}+{\sin{{2}}}{x}}} , if \displaystyle{14}{x}={3}\pi ?
P dilip_k Jan 16, 2018\displaystyle\frac{{{\cos{{5}}}{x}+{\cos{{4}}}{x}}}{{{\sin{{3}}}{x}+{\sin{{2}}}{x}}}\displaystyle=\frac{{{2}{\cos{{\left(\frac{{{9}{x}}}{{2}}\right)}}}{\cos{{\left(\frac{{x}}{{2}}\right)}}}}}{{{2}{\sin{{\left(\frac{{{5}{x}}}{{2}}\right)}}}{\cos{{\left(\frac{{x}}{{2}}\right)}}}}} ...
\displaystyle{x}=\frac{\pi}{{6}},\frac{{{5}\pi}}{{6}}or \displaystyle{x}={30}^{\circ},{150}^{\circ} Explanation:From sum and difference formulasWe have \displaystyle{\sin{{\left({x}+{y}\right)}}}={\sin{{x}}}{\cos{{y}}}+{\cos{{x}}}{\sin{{y}}} ...

Xem thêm: Phát Triển Hệ Sinh Thái Nông Nghiệp Là Gì, Nông Nghiệp Sinh Thái


Writing x = 2y, we obtain \begin{align} \frac{1-\cos x + k\sin x}{\sin x + k(1+\cos x)} &= \frac{(1-\cos (2y)) + k\sin (2y)}{\sin(2y) + k(1+\cos(2y))}\\ &= \frac{2\sin^2 y + 2k\sin y\cos y}{2\sin y\cos y + 2k\cos^2 y}\\ &= \frac{\sin y}{\cos y}\cdot\frac{\sin y+k\cos y}{\sin y + k\cos y}\\ &= \tan y \end{align} ...
More Items
*
*

*
*
*

\left< \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right> \left< \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right>
*
*

*
*

EnglishDeutschEspañolFrançaisItalianoPortuguêsРусский简体中文繁體中文Bahasa MelayuBahasa Indonesiaالعربية日本語TürkçePolskiעבריתČeštinaNederlandsMagyar Nyelv한국어SlovenčinaไทยελληνικάRomânăTiếng Việtहिन्दीঅসমীয়াবাংলাગુજરાતીಕನ್ನಡकोंकणीമലയാളംमराठीଓଡ଼ିଆਪੰਜਾਬੀதமிழ்తెలుగు