The number of solutions x of the equation sin | vedclass

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

Question

(c)

Given,

$\sin \left(x+x^2\right)-\sin \left(x^2\right)=\sin x$

$\Rightarrow \quad \sin \left(x+x^2\right)=\sin \left(x^2\right)+\sin x$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(c)

Given,

$\sin \left(x+x^2\right)-\sin \left(x^2\right)=\sin x$

$\Rightarrow \quad \sin \left(x+x^2\right)=\sin \left(x^2\right)+\sin x$

$\Rightarrow 2 \sin \left(\frac{x+x^2}{2}\right) \cos \left(\frac{x+x^2}{2}\right)$

$=2 \sin \left(\frac{x^2+x}{2}\right) \cos \left(\frac{x^2-x}{2}\right)$

$\sin \left(\frac{x+x^2}{2}\right)=0$

or $\cos \left(\frac{x^2+x}{2}\right)-\cos \left(\frac{x^2-x}{2}\right)=0$

$\frac{x^2+x}{2}=x \pi$ or $2 \sin \frac{x^2}{2} \sin \frac{x}{2}=0$

$\Rightarrow \frac{x^2+x}{2}=0, \pi, 2 \pi$ or $\frac{x^2}{2}=0, \pi, \frac{n}{2}=0, \pi, 2 \pi$

$\Rightarrow x^2+x=2 \pi$ or $x^2=2 \pi$

$\Rightarrow x^2+x-2 \pi=0$ or $x=\sqrt{2 \pi}$

$x=\frac{-1 \pm \sqrt{1+8 \pi}}{2}$ or $2<\sqrt{2} \pi<3$

$\sqrt{1+8 \pi} \approx \sqrt{25.14}$

$x=\frac{5.2-1-4.2-2.1}{2}$

$\because$ 'Total numbers of solution lies between

$(2,3)=2$

The number of solutions $x$ of the equation $\sin \left(x+x^2\right)-\sin \left(x^2\right)=\sin x$ in the interval $[2,3]$ is

Similar Questions

If sum of all the solutions of the equation $8\cos x \cdot \left( {\cos \left( {\frac{\pi }{6} + x} \right) \cdot \cos \left( {\frac{\pi }{6} - x} \right) - \frac{1}{2}} \right) = 1$ in $\left[ {0,\pi } \right]$ is $k\pi $then $k$ is equal to :

If $\cos \theta = - \frac{1}{{\sqrt 2 }}$ and $\tan \theta = 1$, then the general value of $\theta $ is

If $\alpha ,\,\beta ,\,\gamma ,\,\delta $ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity $k$ , then the value of $4\sin \frac{\alpha }{2} + 3\sin \frac{\beta }{2} + 2\sin \frac{\gamma }{2} + \sin \frac{\delta }{2}$ is equal to

Prove that

$\cos 2 x \cos \frac{x}{2}-\cos 3 x \cos \frac{9 x}{2}=\sin 5 x \sin \frac{5 x}{2}$

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