جواب كلی معادلۀ $\operatorname{sinx}\operatorname{cosx}-\frac{۱}{۱+{{\tan }^{۲}}x}=\cos \frac{۴\pi }{۳}$ كدام است؟
$\operatorname{sinx}\operatorname{cosx}-\frac{1}{1+{{\tan }^{2}}x}=\cos \frac{4\pi }{3}\Rightarrow \frac{1}{2}\sin 2x-{{\cos }^{2}}x=-\frac{1}{2}\Rightarrow \sin 2x-2{{\cos }^{2}}x=-1$ $\Rightarrow \sin 2x=2{{\cos }^{2}}x-1\Rightarrow \sin 2x=\cos 2x\Rightarrow tan2x=1\Rightarrow 2x=k\pi +\frac{\pi }{4}\Rightarrow x=\frac{k\pi }{2}+\frac{\pi }{8}$