If 3 sin theta + 4 cos theta = 3 and theta ne | vedclass

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

Question

(D) Given $3 \sin 	heta + 4 \cos 	heta = 3$. Since $	heta 
eq (2n + 1) \frac{\pi}{2}$,$\cos 	heta 
eq 0$,so we can divide by $\cos 	heta$ to ge

Vedclass · Accepted Answer

$-\frac{336}{625}$

Vedclass · Answer

(D) Given $3 \sin 	heta + 4 \cos 	heta = 3$. Since $	heta 
eq (2n + 1) \frac{\pi}{2}$,$\cos 	heta 
eq 0$,so we can divide by $\cos 	heta$ to get $3 	an 	heta + 4 = 3 \sec 	heta$. Squaring both sides: $(3 	an 	heta + 4)^2 = 9 \sec^2 	heta$. $9 	an^2 	heta + 24 	an 	heta + 16 = 9(1 + 	an^2 	heta)$. $9 	an^2 	heta + 24 	an 	heta + 16 = 9 + 9 	an^2 	heta$. $24 	an 	heta = -7 \implies 	an 	heta = -\frac{7}{24}$. We know $\sin 2 	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$. Substituting $	an 	heta = -\frac{7}{24}$: $\sin 2 	heta = \frac{2(-7/24)}{1 + (-7/24)^2} = \frac{-7/12}{1 + 49/576} = \frac{-7/12}{625/576} = -\frac{7}{12} 	imes \frac{576}{625} = -\frac{7 	imes 48}{625} = -\frac{336}{625}$.

If $3 \sin \theta + 4 \cos \theta = 3$ and $\theta \neq (2n + 1) \frac{\pi}{2}$,then $\sin 2 \theta = $

Similar Questions

$\tan \frac{\pi}{5}+2 \tan \frac{2 \pi}{5}+4 \cot \frac{4 \pi}{5}$ is equal to

The number of real solutions of the equation $(\sin x - x)(\cos x - x^2) = 0$ is

The value of the $\sin 1^{\circ} + \sin 2^{\circ} + \ldots + \sin 359^{\circ}$ is equal to

If $\alpha \in (0, \pi/2)$,then $\sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}} = \dots$

If $\cos x + \sin x = \frac{1}{2}$ and $0 < x < \pi$,then $\tan x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module