The equation 3 cos x + 4 sin x = 6 has | vedclass

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

Question

(D) The given equation is $3 \cos x + 4 \sin x = 6$. We know that the range of the expression $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2

Vedclass · Accepted Answer

No solution

Vedclass · Answer

(D) The given equation is $3 \cos x + 4 \sin x = 6$. We know that the range of the expression $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$. Here,$a = 3$ and $b = 4$. So,the range is $[-\sqrt{3^2 + 4^2}, \sqrt{3^2 + 4^2}] = [-\sqrt{9 + 16}, \sqrt{9 + 16}] = [-5, 5]$. Since the value $6$ lies outside the range $[-5, 5]$,the equation $3 \cos x + 4 \sin x = 6$ has no solution.

The equation $3 \cos x + 4 \sin x = 6$ has

Similar Questions

If $\cos 2\theta + 3\cos \theta = 0$,then the general value of $\theta$ is

The solution of the equation $4\cos^2 x + 6\sin^2 x = 5$ is:

The number of solutions of the trigonometric equation $1+\cos x \cdot \cos 5 x=\sin ^2 x$ in $[0, 2 \pi]$ is

The most general value of $\theta$ which satisfies both the equations $\tan \theta = -1$ and $\cos \theta = \frac{1}{\sqrt{2}}$ is

Find the general solution of the equation $\sin 2x + \cos x = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module