Giải bài 1.26 trang 37 - SBT Đại số và Giải tích lớp 11
\(\begin{aligned} & a)\,3{{\cos }^{2}}x-2\sin x+2=0 \\ & \Leftrightarrow 3-3{{\sin }^{2}}x-2\sin x+2=0 \\ & \Leftrightarrow 3{{\sin }^{2}}x+2\sin x-5=0 \\ & \Leftrightarrow \left[ \begin{aligned} & \sin x=1 \\ & \sin x=-\dfrac{5}{3}\,\,\left( \text{loại} \right) \\ \end{aligned} \right. \\ & \Leftrightarrow x=\dfrac{\pi }{2}+k2\pi \,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned} \)
\(\begin{aligned} & b)\,5{{\sin }^{2}}x+3\cos x+3=0 \\ & \Leftrightarrow 5-5{{\cos }^{2}}x+3\cos x+3=0 \\ & \Leftrightarrow 5{{\cos }^{2}}x-3\cos x-8=0 \\ & \Leftrightarrow \left[ \begin{aligned} & \cos x=-1 \\ & \cos x=\dfrac{8}{5}\,\,\left( \text{loại} \right) \\ \end{aligned} \right. \\ & \Leftrightarrow x=\pi +k2\pi \,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned} \)
\(\begin{aligned} & c)\,{{\sin }^{6}}x+{{\cos }^{6}}x=4{{\cos }^{2}}2x \\ & \Leftrightarrow {{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{3}}-3{{\sin }^{2}}x{{\cos }^{2}}x\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)=4{{\cos }^{2}}2x \\ & \Leftrightarrow 1-\dfrac{3}{4}{{\sin }^{2}}2x=4-4{{\sin }^{2}}2x \\ & \Leftrightarrow {{\sin }^{2}}2x=\dfrac{12}{13} \\ & \Leftrightarrow \dfrac{1-\cos 4x}{2}=\dfrac{12}{13} \\ & \Leftrightarrow \cos 4x=-\dfrac{11}{13} \\ & \Leftrightarrow 4x=\pm \arccos \left( -\dfrac{11}{13} \right)+k2\pi \\ & \Leftrightarrow x=\pm \dfrac{1}{4}\arccos \left( -\dfrac{11}{13} \right)+k\dfrac{\pi}{2} \,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned}\)
\(\begin{aligned} & d)-\dfrac{1}{4}+{{\sin }^{2}}x={{\cos }^{4}}x \\ & \Leftrightarrow -\dfrac{1}{4}+\dfrac{1-\cos 2x}{2}={{\left( \dfrac{1+\cos 2x}{2} \right)}^{2}} \\ & \Leftrightarrow -1+2-2\cos 2x=1+2\cos 2x+{{\cos }^{2}}2x \\ & \Leftrightarrow {{\cos }^{2}}2x+4\cos 2x=0 \\ & \Leftrightarrow \left[ \begin{aligned} & \cos 2x=0 \\ & \cos 2x=-4\,\,\left( \text{loại} \right) \\ \end{aligned} \right. \\ & \Leftrightarrow 2x=\dfrac{\pi }{2}+k\pi \\ & \Leftrightarrow x=\dfrac{\pi }{4}+k\dfrac{\pi}{2} \,\,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned} \)