$\cos x+\sqrt{2}=-\cos x \\ 2\cos x = -\sqrt 2 \\ \cos x = -\dfrac{\sqrt 2}{2}\\ \\ \Rightarrow x = \pm \arccos\Big(-\dfrac{\sqrt 2}{2}\Big)+2k\pi,\quad x\in \mathbb{Z}\\ \Rightarrow x =\pm\dfrac{3\pi}{4}+2k\pi\\ \\ \\k = -1 \Rightarrow x < 0\\\\ k=0 \Rightarrow x 2\pi$

3 0

3 years ago

Read 2 more answers

Please can someone help me on these? solve this equation by factorising: 2x^2 + 5x + 2 = 0

slavikrds [6]

$2x^2 + 5x + 2 = 0\\
2x^2+4x+x+2=0\\
2x(x+2)+1(x+2)=0\\
(2x+1)(x+2)=0\\
x=-\frac{1}{2} \vee x=-2$

$x^2-4x+1=0\\
x^2-4x+4-3=0\\
(x-2)^2=3\\
x-2 =\sqrt3 \vee x-2=-\sqrt3\\
x=2+\sqrt3 \vee x=2-\sqrt3$

7 0

3 years ago

Plz help fast will mark brainliest!!!

quester [9]

Answer:

well the answer is 0.185

Step-by-step explanation:

5 0

2 years ago

I will give brainliest to person who answers correctly!

Crank

Answer:

41 is the answer .......

4 0

3 years ago