Bài tập 1.26 trang 37 SBT Toán 11

a) 3cos²x−2sinx+2 = 0

⇔ 3(1−sin²x)−2sin⁡x+2 = 0

⇔3sin²x+2sinx−5 = 0

⇔(sinx−1).(3sinx+5) = 0

\( \Leftrightarrow \left[ \begin{array}{l}
\sin x - 1 = 0\\
3\sin x + 5 = 0
\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x =  - \frac{5}{3}\,\,(vn)
\end{array} \right.\)

\( \Leftrightarrow x = \frac{\pi }{2} + k2\pi ,k \in Z\)

b) 5sin²x+3cosx+3 = 0

⇔ 5(1−cos²x)+3cosx+3 = 0

⇔ 5cos²x−3cosx−8 = 0

⇔ (cosx+1)(5cosx−8) = 0

\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\cos x + 1 = 0}\\
{5\cos x - 8 = 0}
\end{array}} \right.\)

\(\left[ \begin{array}{l}
\cos x =  - 1\\
\cos x = \frac{8}{5}\,\,\left( {vn} \right)
\end{array} \right.\)

\( \Leftrightarrow x = (2k + 1)\pi ,k \in Z\)

c) sin⁶x+cos⁶x = 4cos²2x

⇔ (sin²x+cos²x)³ - 3sin²xcos²x(sin²x+cos²x) = 4cos²2x

\(\begin{array}{l}
 \Leftrightarrow 1 - \frac{3}{4}{\sin ^2}2x = 4{\cos ^2}2x\\
 \Leftrightarrow 1 - \frac{3}{4}(1 - {\cos ^2}2x) = 4{\cos ^2}2x\\
 \Leftrightarrow \frac{{13}}{4}{\cos ^2}2x = \frac{1}{4}\\
 \Leftrightarrow 13\left( {\frac{{1 + \cos 4x}}{2}} \right) = 1\\
 \Leftrightarrow 1 + \cos 4x = \frac{2}{{13}}\\
 \Leftrightarrow \cos 4x =  - \frac{{11}}{{13}}\\
 \Leftrightarrow 4x =  \pm \arccos \left( { - \frac{{11}}{{13}}} \right) + k2\pi ,k \in Z\\
 \Leftrightarrow x =  \pm \frac{1}{4}\arccos \left( { - \frac{{11}}{{13}}} \right) + k\frac{\pi }{2},k \in Z
\end{array}\)

d) \( - \frac{1}{4} + {\sin ^2}x = {\cos ^4}x\)

\(\begin{array}{l}
 \Leftrightarrow  - \frac{1}{4} + \frac{{1 - \cos 2x}}{2} = {\left( {\frac{{1 + \cos 2x}}{2}} \right)^2}\\
 \Leftrightarrow  - 1 + 2 - 2\cos 2x = 1 + 2\cos 2x + {\cos ^2}2x\\
 \Leftrightarrow {\cos ^2}2x + 4\cos 2x = 0\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\cos 2x = 0}\\
{\cos 2x =  - 4\,\,\left( {vn} \right)}
\end{array}} \right.\\
 \Leftrightarrow 2x = \frac{\pi }{2} + k\pi ,k \in Z\\
 \Leftrightarrow x = \frac{\pi }{4} + k\frac{\pi }{2},k \in Z
\end{array}\)

-- Mod Toán 11 HỌC247