সমীকরণ সমাধান
cos θ - cos 7θ = sin 4θ সমীকরণে θ এর মান -
nπ2+π3 \frac{n π}{2} + \frac{π}{3} 2nπ+3π
nπ3+π16 \frac{n π}{3} + \frac{π}{16} 3nπ+16π
nπ3+(−1)nπ18 \frac{n π}{3} + \left ( - 1 \right )^{n} \frac{π}{18} 3nπ+(−1)n18π
nπ3+(−1)nπ6 \frac{n π}{3} + \left ( - 1 \right )^{n} \frac{π}{6} 3nπ+(−1)n6π
cosθ−cos7θ=sin4θ⇒2sin(θ+7θ2)sin(7θ−θ2)=sin4θ⇒2sin4θsin3θ−sin4θ=0\cos\theta-\cos7\theta=\sin4\theta\\ \Rightarrow2\sin\left(\frac{\theta+7\theta}{2}\right)\sin\left(\frac{7\theta-\theta}{2}\right)=\sin4\theta\\\Rightarrow2\sin4\theta\sin3\theta-\sin4\theta=0cosθ−cos7θ=sin4θ⇒2sin(2θ+7θ)sin(27θ−θ)=sin4θ⇒2sin4θsin3θ−sin4θ=0
⇒sin4θ(2sin3θ−1)=0\Rightarrow\sin4\theta\left(2\sin3\theta-1\right)=0⇒sin4θ(2sin3θ−1)=0
∴sin4θ=0 or, 2sin3θ−1=0\therefore\sin4\theta=0\ \ or,\ \ 2\sin3\theta-1=0∴sin4θ=0 or, 2sin3θ−1=0
θ=nπ4 or, nπ3+(−1)nπ18\theta=\frac{n\pi}{4\ }\ \ or,\ \frac{n\pi}{3}+\frac{\left(-1\right)^n\pi}{18}θ=4 nπ or, 3nπ+18(−1)nπ
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 -
