Share
Ai có thể giúp mình giải bìa này ko ak mình đang cần gấp ☹ ai giải đc mình cho câu tlhn luôn ak
Question
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Answers ( )
`a) 3cos² x + 2cos x – 1 = 0`
`<=>` \(\left[ \begin{array}{l}cos x = \dfrac{1}{3}\\cos x = -1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = ±arccos \dfrac{1}{3} + k2π\\x = π + k2π\end{array} \right.\) `(k ∈ ZZ)`
`b) 2sin² x + 5cos x + 1 = 0`
`<=> 2 – 2cos² x + 5cos x + 1 = 0`
`<=> -2cos² x + 5cos x + 3 = 0`
`<=>` \(\left[ \begin{array}{l}cos x = 3 (l)\\cos x = -\dfrac{1}{2}\end{array} \right.\)
`<=> x = ±(2π)/3 + k2π`
`c) cos² x – 4cos x + 5/2 = 0`
`<=> 2cos² x – 8cos x + 5 = 0`
`<=>` \(\left[ \begin{array}{l}cos x = \dfrac{4 + \sqrt{6}}{2} (l)\\x = \dfrac{4 – \sqrt{6}}{2}\end{array} \right.\)
`<=> x = ±arccos ((4 – \sqrt{6})/2) + k2π` `(k ∈ ZZ)`
`d) cos² x – cos x – 2 = 0`
`<=>` \(\left[ \begin{array}{l}cos x = 2 (l)\\cos x = -1\end{array} \right.\)
`<=> x = π + k2π` `(k ∈ ZZ)`
`e) 16 – 15sin² x – 8cos x = 0`
`<=> 16 – 15 + 15cos² x – 8cos x = 0`
`<=> 15cos² x – 8cos x + 1 = 0`
`<=>` \(\left[ \begin{array}{l}cos x = \dfrac{1}{3}\\cos x = \dfrac{1}{5}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = ±arccos \dfrac{1}{3} + k2π\\x = arccos \dfrac{1}{5} + k2π\end{array} \right.\) `(k ∈ ZZ)`