The set of angles btween 0 & 2pi satisfy | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$4\, cos^2	heta - 2 \,cos	heta - 1 = 0$

$cos	heta =$$\frac{{2\sqrt 2 \, \pm \,\sqrt {8 + 16} }}{8}$  $=$$\frac{{\sqrt 2 \, \pm \,\sqrt

Vedclass · Accepted Answer

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

Vedclass · Answer

$4\, cos^2	heta - 2 \,cos	heta - 1 = 0$

$cos	heta =$$\frac{{2\sqrt 2 \, \pm \,\sqrt {8 + 16} }}{8}$  $=$$\frac{{\sqrt 2 \, \pm \,\sqrt 6 }}{4}$

$cos	heta =$ $\frac{{\sqrt 6  + \sqrt 2 }}{4}$ $ \Rightarrow \,\,	heta \,\, = \,\,\frac{\pi }{{12}}\,;\,2\pi \, - \frac{\pi }{{12}}\, = \,\frac{{23\pi }}{{12}}$

$cos	heta =$ $ - \,\frac{{\sqrt 6  - \sqrt 2 }}{4}$

$cos	heta = cos(\pi -5\pi /12) ; cos(\pi +5\pi /12)$

$	heta = 7\pi /12 ; 17\pi /12$

The set of angles btween $0$ & $2\pi $ satisfying the equation $4\, cos^2 \, \theta - 2 \sqrt 2 \, cos \,\theta - 1 = 0$ is

Similar Questions

If the equation $\cos ^{4} \theta+\sin ^{4} \theta+\lambda=0$ has real solutions for $\theta,$ then $\lambda$ lies in the interval

If $sin^4\,\,\alpha + 4\,cos^4\,\,\beta + 2 = 4\sqrt 2\,\,sin\,\alpha \,cos\,\beta ;$ $\alpha \,,\,\beta \, \in \,[0,\pi ],$ then $cos( \alpha + \beta)$ is equal to

The general solution of the trigonometric equation $tan\, x + tan \,2x + tan\, 3x = tan \,x · tan\, 2x · tan \,3x$ is

If $\cos \theta + \cos 7\theta + \cos 3\theta + \cos 5\theta = 0$, then $\theta $

The number of solutions of the given equation $\tan \theta + \sec \theta = \sqrt 3 ,$ where $0 < \theta < 2\pi $ is