tan \(x=3\Rightarrow tan^2x+1=\dfrac{1}{cos^2x}=10\Rightarrow cos^2x=\dfrac{1}{10}\Rightarrow sin^2x=\dfrac{9}{10}\)
\(P=cos2x+5sin^2\left(-x\right)=cos^2x-sin^2x+5sin^2x=cos^2x+4sin^2x=\dfrac{1}{10}+4.\dfrac{9}{10}=\dfrac{37}{10}\)
$\bullet$`1 + tan^2 x = 1 / [ cos^2 x] => 1 + 3^2 = 1 / [ cos^2 x ]`
`=> cos^2 x = 1 / 10`
$\bullet$`sin^2x + cos^2 x = 1 => sin^2 x + 1 / 10 = 1`
`=> sin^2 x = 9 / 10`
Có: `P = cos 2x + 5 sin^2 (-x) = 1 - 2 sin^2 x + 5 sin^2 x = 1 + 3 sin^2 x`
`=> P = 1 + 3 . 9 / 10 = 37 / 10`
