2cos^2(x+pi/3)+5sin(x+pi/3)-4=0 Cos2x-4cosx+5/2=0 Sin^4x+cos^4x=cos2x Cos^4x+sin^4x=sin2x-1/2 24/09/2021 Bởi Camila 2cos^2(x+pi/3)+5sin(x+pi/3)-4=0 Cos2x-4cosx+5/2=0 Sin^4x+cos^4x=cos2x Cos^4x+sin^4x=sin2x-1/2
3) ${\sin}^4x+{\cos}^4x=\cos 2x$ $\Rightarrow (\dfrac{1+\cos 2x}{2})^2+(\dfrac{1-\cos 2x}{2})^2=\cos 2x$ $\Rightarrow \dfrac{1+2\cos2x+{\cos}^22x}{4}+\dfrac{1-2\cos2x+{\cos}^22x}{4}=\cos 2x$ $\Rightarrow 1+\dfrac{{\cos}^22x}{2}=\cos 2x$ $\Rightarrow \cos 2x=1$ $\Rightarrow 2x=\dfrac{\pi}{2}+k\pi$ $\Rightarrow x=\dfrac{\pi}{4}+k\dfrac{\pi}{2},(k\in\mathbb Z)$ 2) $\cos 2x-4\cos x+\dfrac{5}{2}=0$ $\Rightarrow 2{\cos}^2x-1-4\cos x+\dfrac{5}{2}=0$ $\Rightarrow \left[ \begin{array}{l} \cos x=\dfrac{3}{2}>1\text{ (loại)} \\ \cos x=1(tm) \end{array} \right .$ $\Rightarrow x=k2\pi,(k\in\mathbb Z)$. Bình luận
Đáp án: Giải thích các bước giải: a) 2cos^2(x + π/3) + 5sin(x+ π/3) – 4 =0 => 2(1 – sin^2)(x + π/3)+5sin(x + π/3) – 4= 0 => 2 – 2sin^2(x+π/3)+5sin(x+π/3) – 4 =0 => -2sin^2(x+π/3) + 5sin(x+π/3) – 2=0 Bạn bấm mode 5 3 rồi làm tiếp nha b) cos2x – 4cosx +5/2 =0 => 2cos^2x – 1 – 4cosx + 5/2 =0 => 2cos^2x – 4cosx + 3/2 =0 Bạn bấm mode 5 3 rồi làm tiếp nha c) Sin^4x + cos^4x = cos2x =>1 – 2sin^2xcos^2x = cos2x => 1- 2 (sin^2 2x)/4 = cos2x => 1 – (sin^2 2x)/2 = cos2x => 2 – sin^2 2x = 2cos2x => 2 – (1 – cos^2 2x) = 2cos2x => cos^2 2x – 2cos2x +1 =0 Bạn bấm mode 5 3 rồi giải tiếp nha Câu d tương tự câu c Bình luận
3) ${\sin}^4x+{\cos}^4x=\cos 2x$
$\Rightarrow (\dfrac{1+\cos 2x}{2})^2+(\dfrac{1-\cos 2x}{2})^2=\cos 2x$
$\Rightarrow \dfrac{1+2\cos2x+{\cos}^22x}{4}+\dfrac{1-2\cos2x+{\cos}^22x}{4}=\cos 2x$
$\Rightarrow 1+\dfrac{{\cos}^22x}{2}=\cos 2x$
$\Rightarrow \cos 2x=1$
$\Rightarrow 2x=\dfrac{\pi}{2}+k\pi$
$\Rightarrow x=\dfrac{\pi}{4}+k\dfrac{\pi}{2},(k\in\mathbb Z)$
2) $\cos 2x-4\cos x+\dfrac{5}{2}=0$
$\Rightarrow 2{\cos}^2x-1-4\cos x+\dfrac{5}{2}=0$
$\Rightarrow \left[ \begin{array}{l} \cos x=\dfrac{3}{2}>1\text{ (loại)} \\ \cos x=1(tm) \end{array} \right .$
$\Rightarrow x=k2\pi,(k\in\mathbb Z)$.
Đáp án:
Giải thích các bước giải:
a) 2cos^2(x + π/3) + 5sin(x+ π/3) – 4 =0
=> 2(1 – sin^2)(x + π/3)+5sin(x + π/3) – 4= 0
=> 2 – 2sin^2(x+π/3)+5sin(x+π/3) – 4 =0
=> -2sin^2(x+π/3) + 5sin(x+π/3) – 2=0
Bạn bấm mode 5 3 rồi làm tiếp nha
b) cos2x – 4cosx +5/2 =0
=> 2cos^2x – 1 – 4cosx + 5/2 =0
=> 2cos^2x – 4cosx + 3/2 =0
Bạn bấm mode 5 3 rồi làm tiếp nha
c) Sin^4x + cos^4x = cos2x
=>1 – 2sin^2xcos^2x = cos2x
=> 1- 2 (sin^2 2x)/4 = cos2x
=> 1 – (sin^2 2x)/2 = cos2x
=> 2 – sin^2 2x = 2cos2x
=> 2 – (1 – cos^2 2x) = 2cos2x
=> cos^2 2x – 2cos2x +1 =0
Bạn bấm mode 5 3 rồi giải tiếp nha
Câu d tương tự câu c