সমীকরণ সমাধান
If cos3x=−1\cos 3x=-1cos3x=−1, where 0o≤x≤360o0^{o} \le x \le 360^{o}0o≤x≤360o, then xxx
60o,180o,300o60^{o},180^{o},300^{o}60o,180o,300o
180o180^{o}180o
60o,180o60^{o},180^{o}60o,180o
180o,300o180^{o},300^{o}180o,300o
cos3x=−1\cos 3 x=-1cos3x=−1
⟹ 3x=2nπ±π=(2n±1)π\implies 3 x=2 n \pi\pm \pi=(2 n \pm 1)\pi⟹3x=2nπ±π=(2n±1)π
⟹ x=(2n±1)π3\implies x=\dfrac{(2 n\pm 1)\pi}{3}⟹x=3(2n±1)π
⟹ x=(2n±1)60∘=±60∘,±180∘,±300∘,±420∘,⋯\implies x=(2 n\pm 1)60^{\circ}=\pm 60^{\circ},\pm 180^{\circ},\pm 300^{\circ},\pm 420^{\circ},\cdots⟹x=(2n±1)60∘=±60∘,±180∘,±300∘,±420∘,⋯
As 0∘≤x≤360∘,x=60∘,180∘,300∘0^{\circ}\le x\le 360^{\circ},x=60^{\circ},180^{\circ},300^{\circ} 0∘≤x≤360∘,x=60∘,180∘,300∘
f(x)=sinx \mathrm{f}(x)=\sin x f(x)=sinx এবং g(x)=cosx g(x)=\cos x g(x)=cosx.
3sec-1(2)=cos-1x হলে x এর মান ত?
cosθ=12 \cos{θ} = \frac{1}{\sqrt{2}} cosθ=21 হলে,θ এর মান কত?
2tan−1(cosx)=tan−1(2cosecx) 2 \tan^{- 1}{\left ( \cos{x} \right )} = \tan^{- 1}{\left ( 2 \cos{e} c x \right )} 2tan−1(cosx)=tan−1(2cosecx) এর সমাধান -
