Пусть y = uv, тогда y' = u'v + uv':
Решим левый интеграл:
cosx = \frac{1-t^2}{1+t^2} => dx = \frac{2}{1+t^2}dt\\ \int \frac{2(1+t^2)}{(1+t^2)(1-t^2)} dt = \int \frac{2}{(1-t)(1+t)}dt = \int ( \frac{1}{1-t} + \frac{1}{1+t})dt = ln(1-t)+ln( 1+t) = ln|1-t^2| = ln|1-tg^2\frac{x}{2}| \\" class="latex-formula" id="TexFormula2" src="https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7Bdx%7D%7Bcosx%7D%3B%5C%5C%20tg%5Cfrac%7Bx%7D%7B2%7D%3Dt%20%3D%3E%20cosx%20%3D%20%5Cfrac%7B1-t%5E2%7D%7B1%2Bt%5E2%7D%20%3D%3E%20dx%20%3D%20%5Cfrac%7B2%7D%7B1%2Bt%5E2%7Ddt%5C%5C%20%20%5Cint%20%5Cfrac%7B2%281%2Bt%5E2%29%7D%7B%281%2Bt%5E2%29%281-t%5E2%29%7D%20dt%20%3D%20%5Cint%20%5Cfrac%7B2%7D%7B%281-t%29%281%2Bt%29%7Ddt%20%3D%20%5Cint%20%28%20%5Cfrac%7B1%7D%7B1-t%7D%20%2B%20%5Cfrac%7B1%7D%7B1%2Bt%7D%29dt%20%3D%20ln%281-t%29%2Bln%28%201%2Bt%29%20%3D%20ln%7C1-t%5E2%7C%20%3D%20ln%7C1-tg%5E2%5Cfrac%7Bx%7D%7B2%7D%7C%20%20%5C%5C" title="\int \frac{dx}{cosx};\\ tg\frac{x}{2}=t => cosx = \frac{1-t^2}{1+t^2} => dx = \frac{2}{1+t^2}dt\\ \int \frac{2(1+t^2)}{(1+t^2)(1-t^2)} dt = \int \frac{2}{(1-t)(1+t)}dt = \int ( \frac{1}{1-t} + \frac{1}{1+t})dt = ln(1-t)+ln( 1+t) = ln|1-t^2| = ln|1-tg^2\frac{x}{2}| \\">
Возвращаемся к исходному:
6sinxcosx+8cos²x-7sin²x-7cos²x=0
7sin²x-6sinxcosx-1=0/cos²x
7tg²x-6tgx-1=0
tgx=t
7t²-6t-1=0
D=36+28=64
t1=(6-8)/14=-1/7⇒tgx=-1/7⇒x=-arctg1/7+πk,k∈z
t2=(6+8)/14=1⇒tgx=1⇒x=π/4+πk,k∈z
2. (cosx-2)/cos(x/2)=2
cos(x/2)≠0⇒x/2≠π/2+πk⇒x≠π+2πk,k∈z
2cos²(x/2)-1-3-2cos(x/2)=0
cos(x/2)=t
2t²-2t-3=0
D=4+24=28
t1=(2-2√7)/4=0,5-0,5√7⇒cos(x/2)=0,5-0,5√7
x/2=+-arccos(0,5-0,5√7)+2πk
x=+-2arccos(0,5-0,5√7)+2πk,k∈z
t2=0,5+0,5√7⇒cos(x/2)=0,5+0,5√7>1 нет решения
3. 1+sin2x ×cosx=sin2x+cosx
(sin2xcosx-sin2x)+(1-cosx)=0
sin2x(cosx-1)-(cosx-1)=0
(cosx-1)(sin2x-1)=0
cosx=1⇒x=2πk,k∈z
sin2x=1⇒2x=π/2+2πk⇒x=π/4+πk,k∈z