If equation 2 {sin ^2}x + frac{{sin 2x}}{2} = k , has atleast one real solution, the

English

Gujarati

Hindi

Trigonometrical Equations

normal

If the equation $2\ {\sin ^2}x + \frac{{\sin 2x}}{2} = k$ , has atleast one real solution, then the sum of all integral values of $k$ is

A

$2$

B

$3$

C

$5$

D

$6$

Solution

$-\cos 2 x+\frac{\sin 2 x}{2}=k-1$

$-\frac{\sqrt{5}}{2} \leq k-1 \leq \frac{\sqrt{5}}{2}$

$ \Rightarrow 1-\frac{\sqrt{5}}{2} \leq k \leq 1+\frac{\sqrt{5}}{2}$

$\therefore \operatorname{sum}=0+1+2=3$

Standard 11

Mathematics

Share

Similar Questions

If $0\, \le \,x\, < \frac{\pi }{2},$ then the number of values of $x$ for which $sin\,x -sin\,2x + sin\,3x=0,$ is

hard

(JEE MAIN-2019)

If $\sec 4\theta – \sec 2\theta = 2$, then the general value of $\theta $ is

medium

(IIT-1963)

The general value $\theta $ is obtained from the equation $\cos 2\theta = \sin \alpha ,$ is

easy

If the solution of the equation $\log _{\cos x} \cot x+4 \log _{\sin x} \tan x=1, x \in\left(0, \frac{\pi}{2}\right), \quad$ is $\sin ^{-1}\left(\frac{\alpha+\sqrt{\beta}}{2}\right)$, where $\alpha, \beta$ are integers, then $\alpha+\beta$ is equal to:

hard

(JEE MAIN-2023)

For which value of $x$ ; $cosx > sinx,$ where $x\, \in \,\,\left( {\frac{\pi }{2}\,,\,\frac{{3\pi }}{2}} \right)$

normal