Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De oplossingenverzameling van sin 2x = 2.sin x is (in radialen) ( k elk geheel getal )	A. { k.π}
B. { k. }
C. { + k.2π }
D. { k. }
E.

The solution set of sin 2x = 2.sin x is (in radians) ( k integer )	A. { k.π }
B. { k. }
C. { + k.2π }
D. { k. }
E.

1^ste manier : (in zestigdelige graden)
sin 2x = 2.sin x
⇔ 2.sinx.cosx − 2.sin x = 0
⇔ 2sinx(cos x − 1) = 0
⇔ sin x = 0 ∨ cos x = 1
⇔ x = k.180° ∨ x = 0° + k.360°
⇔ x = k.180°
2^de manier : (in zestigdelige graden)
$\sin2x=2.\sin x\\\Leftrightarrow\;\frac{2\tan x}{1+\tan^2x}=2\tan x.\cos x\\\Leftrightarrow\;\frac{\tan x}{1+\tan^2x}=\tan x.\frac{1}{\sqrt{1+\tan^2x}}\\\Leftrightarrow\;\frac{\tan x}{1+\tan^2x}-\tan x.\frac{1}{\sqrt{1+\tan^2x}}=0\\\tan x\left(\frac{1}{1+\tan^2x}-\frac{1}{\sqrt{1+\tan^2x}}\right)=0\\\Leftrightarrow\;\tan x=0\;\;\vee\;\;\frac{1}{1+\tan^2x}=\frac{1}{\sqrt{1+\tan^2x}}\\\Leftrightarrow\;\tan x=0\;\;\vee\;\;1+\tan^2x=\sqrt{1+\tan^2x}\\\Leftrightarrow\;\tan x=0\;\;\vee\;\;\tan^2x=0\\\Leftrightarrow\;\tan x=0\\\Leftrightarrow\;x=k.180^\circ$