If $\cos \theta = - \frac{1}{{\sqrt 2 }}$ and $\tan \theta = 1$, then the general value of $\theta $ is
If $\cos \theta = - \frac{1}{{\sqrt 2 }}$ and $\tan \theta = 1$, then the general value of $\theta $ is
- A
$2n\pi + \frac{\pi }{4}$
- B
$(2n + 1)\,\pi + \frac{\pi }{4}$
- C
$n\pi + \frac{\pi }{4}$
- D
$n\pi \pm \frac{\pi }{4}$
Similar Questions
If $x = \frac{{n\pi }}{2}$ , satisfies the equation $sin\, \frac{x}{2}- cos \frac{x}{2} = 1$ $- sin\, x$ & the inequality $\left| {\frac{x}{2}\,\, - \,\,\frac{\pi }{2}} \right|\,\, \le \,\,\frac{{3\pi }}{4}$, then:
If $x = \frac{{n\pi }}{2}$ , satisfies the equation $sin\, \frac{x}{2}- cos \frac{x}{2} = 1$ $- sin\, x$ & the inequality $\left| {\frac{x}{2}\,\, - \,\,\frac{\pi }{2}} \right|\,\, \le \,\,\frac{{3\pi }}{4}$, then:
If $\sec x\cos 5x + 1 = 0$, where $0 < x < 2\pi $, then $x =$
If $\sec x\cos 5x + 1 = 0$, where $0 < x < 2\pi $, then $x =$
- [IIT 1963]
The equation ${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$ is solvable for
The equation ${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$ is solvable for
One root of the equation $\cos x - x + \frac{1}{2} = 0$ lies in the interval
One root of the equation $\cos x - x + \frac{1}{2} = 0$ lies in the interval
The roots of the equation $1 - \cos \theta = \sin \theta .\sin \frac{\theta }{2}$ is
The roots of the equation $1 - \cos \theta = \sin \theta .\sin \frac{\theta }{2}$ is