4, cos^2 , θ - 2 √ 2 , cos ,θ - 1 = 0 સમીકરણને સંતોષતી

Hindi

Trigonometrical Equations

normal

$4\, cos^2 \, \theta - 2 \sqrt 2 \, cos \,\theta - 1 = 0$ સમીકરણને સંતોષતી $0$ & $2\pi $ ની વચ્ચેની કિમત .............. છે

$\left\{ {\frac{\pi }{{12}}\,\,,\,\,\frac{{5\pi }}{{12}}\,\,,\,\,\frac{{19\pi }}{{12}}\,\,,\,\,\frac{{23\pi }}{{12}}} \right\}$

$\left\{ {\frac{\pi }{{12}}\,\,,\,\,\frac{{7\pi }}{{12}}\,\,,\,\,\frac{{17\pi }}{{12}}\,\,,\,\,\frac{{23\pi }}{{12}}} \right\}$

$\left\{ {\,\,\frac{{5\pi }}{{12}}\,\,,\,\,\frac{{13\pi }}{{12}}\,\,,\,\,\frac{{19\pi }}{{12}}} \right\}$

$\left\{ {\frac{\pi }{{12}}\,\,,\,\,\frac{{7\pi }}{{12}}\,\,,\,\,\frac{{19\pi }}{{12}}\,\,,\,\,\frac{{23\pi }}{{12}}} \right\}$

Solution

$4\, cos^2\theta – 2 \,cos\theta – 1 = 0$

$cos\theta =$$\frac{{2\sqrt 2 \, \pm \,\sqrt {8 + 16} }}{8}$ $=$$\frac{{\sqrt 2 \, \pm \,\sqrt 6 }}{4}$

$cos\theta =$ $\frac{{\sqrt 6 + \sqrt 2 }}{4}$ $ \Rightarrow \,\,\theta \,\, = \,\,\frac{\pi }{{12}}\,;\,2\pi \, – \frac{\pi }{{12}}\, = \,\frac{{23\pi }}{{12}}$

$cos\theta =$ $ – \,\frac{{\sqrt 6 – \sqrt 2 }}{4}$

$cos\theta = cos(\pi -5\pi /12) ; cos(\pi +5\pi /12)$

$\theta = 7\pi /12 ; 17\pi /12$

Standard 11

Mathematics

જો $0 \le x \le \pi $ અને ${81^{{{\sin }^2}x}} + {81^{{{\cos }^2}x}} = 30$, તો $x =$

hard

(JEE MAIN-2021)

$\cos x=\frac{1}{2}$ ઉકેલો.

easy

અહી $S={\theta \in\left(0, \frac{\pi}{2}\right): \sum_{m=1}^{9}}$

$\sec \left(\theta+(m-1) \frac{\pi}{6}\right) \sec \left(\theta+\frac{m \pi}{6}\right)=-\frac{8}{\sqrt{3}}$ હોય તો . . .

hard

(JEE MAIN-2022)

સમીકરણ $\left| {\,\begin{array}{*{20}{c}}{\cos \theta }&{\sin \theta }&{\cos \theta }\\{ – \sin \theta }&{\cos \theta }&{\sin \theta }\\{ – \cos \theta }&{ – \sin \theta }&{\cos \theta }\end{array}\,} \right| = 0$ નો ઉકેલ મેળવો.

medium

જો $\tan (\pi \cos \theta ) = \cot (\pi \sin \theta )$, તો $\sin \left( {\theta + \frac{\pi }{4}} \right) = . . . .$