Bài tập 1.29 trang 38 SBT Toán 11

a) Ta có 2cosx−sinx = 2

\( \Leftrightarrow \frac{2}{{\sqrt 5 }}\cos x - \frac{1}{{\sqrt 5 }}\sin x = \frac{2}{{\sqrt 5 }}\)

Ký hiệu α là góc mà \(\cos \alpha  = \frac{2}{{\sqrt 5 }}\) và \(\sin \alpha  =  - \frac{1}{{\sqrt 5 }}\)

Ta thu được phương trình

cosα.cosx+sinα.sinx = cosα

⇔ cos(x−α) = cosα

\(\begin{array}{l}
 \Leftrightarrow x - \alpha  =  \pm \alpha  + k2\pi ,k \in Z\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\alpha  + k2\pi ,k \in Z\\
x = k2\pi ,k \in Z
\end{array} \right.
\end{array}\)

b) Ta có sin5x+cos5x = −1

\( \Leftrightarrow \frac{1}{{\sqrt 2 }}\cos 5x + \frac{1}{{\sqrt 2 }}\sin 5x =  - \frac{1}{{\sqrt 2 }}\)

Trong đó \(\cos \frac{\pi }{4} = \frac{1}{{\sqrt 2 }},sin\frac{\pi }{4} = \frac{1}{{\sqrt 2 }}\) và \(\sin \left( { - \frac{\pi }{4}} \right) =  - \frac{1}{{\sqrt 2 }}\)

Ta thu được phương trình

\(\begin{array}{l}
\cos \frac{\pi }{4}\sin 5x + \sin \frac{\pi }{4}\cos 5x = \sin \left( { - \frac{\pi }{4}} \right)\\
 \Leftrightarrow \sin (5x + \frac{\pi }{4}) = \sin \left( { - \frac{\pi }{4}} \right)\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{5x + \frac{\pi }{4} =  - \frac{\pi }{4} + k2\pi ,k \in Z}\\
{5x + \frac{\pi }{4} = \pi  - \left( { - \frac{\pi }{4}} \right) + k2\pi ,k \in Z}
\end{array}} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \frac{\pi }{{10}} + k\frac{{2\pi }}{5},k \in Z\\
x = \frac{\pi }{5} + k\frac{{2\pi }}{5},k \in Z
\end{array} \right.
\end{array}\)

c) Ta có 8cos⁴x−4cos2x+sin4x−4 = 0

\(\begin{array}{l}
 \Leftrightarrow 8{\left( {\frac{{1 + \cos 2x}}{2}} \right)^2} - 4\cos 2x + \sin 4x - 4 = 0\\
 \Leftrightarrow 2(1 + 2\cos 2x + {\cos ^2}2x) - 4\cos 2x + \sin 4x - 4 = 0\\
 \Leftrightarrow 2{\cos ^2}2x + \sin 4x - 2 = 0\\
 \Leftrightarrow 1 + \cos 4x + \sin 4x - 2 = 0\\
 \Leftrightarrow \cos 4x + \sin 4x = 1\\
 \Leftrightarrow \frac{1}{{\sqrt 2 }}\cos 4x + \frac{1}{{\sqrt 2 }}\sin 4x = \sin \frac{\pi }{4}\\
 \Leftrightarrow \sin \frac{\pi }{4}\cos 4x + \cos \frac{\pi }{4}\sin 4x = \sin \frac{\pi }{4}\\
 \Leftrightarrow \sin \left( {4x + \frac{\pi }{4}} \right) = \sin \frac{\pi }{4}\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{4x + \frac{\pi }{4} = \frac{\pi }{4} + k2\pi ,k \in Z}\\
{4x + \frac{\pi }{4} = \pi  - \frac{\pi }{4} + k2\pi ,k \in Z}
\end{array}} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\frac{\pi }{2},k \in Z\\
x = \frac{\pi }{8} + k\frac{\pi }{2},k \in Z
\end{array} \right.
\end{array}\)

d) Ta có \({\sin ^6}x + {\cos ^6}x + \frac{1}{2}\sin 4x = 0\)$\mathrm{}$

\(\begin{array}{l}
 \Leftrightarrow {({\sin ^2}x + {\cos ^2}x)^3} - 3{\sin ^2}x{\cos ^2}x({\sin ^2}x + {\cos ^2}x) + \frac{1}{2}\sin 4x = 0\\
 \Leftrightarrow 1 - 3{\sin ^2}x{\cos ^2}x + \frac{1}{2}\sin 4x = 0\\
 \Leftrightarrow 1 - 3{\left( {\frac{{\sin 2x}}{2}} \right)^2} + \frac{1}{2}\sin 4x = 0\\
 \Leftrightarrow 1 - \frac{3}{4}{\sin ^2}2x + \frac{1}{2}\sin 4x = 0\\
 \Leftrightarrow 1 - \frac{3}{4}.\frac{{1 - \cos 4x}}{2} + \frac{1}{2}\sin 4x = 0\\
 \Leftrightarrow 8 - 3 + 3\cos 4x + 4\sin 4x = 0\\
 \Leftrightarrow 3\cos 4x + 4\sin 4x =  - 5\\
 \Leftrightarrow \frac{3}{5}\cos 4x + \frac{4}{5}\sin 4x =  - 1
\end{array}\)

Đặt \(\frac{3}{5} = \sin \alpha ,\frac{4}{5} = \cos \alpha \) ta được

\(\begin{array}{l}
\sin \alpha \cos 4x + \cos \alpha \sin 4x =  - 1\\
 \Leftrightarrow \sin (4x + \alpha ) =  - 1\\
 \Leftrightarrow 4x + \alpha  = \frac{{3\pi }}{2} + k2\pi ,k \in Z\\
 \Leftrightarrow x = \frac{{3\pi }}{8} - \frac{\alpha }{4} + k\frac{\pi }{2},k \in Z
\end{array}\)

-- Mod Toán 11 HỌC247