The general solution of the equation sqrt{6-5 | vedclass

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

Question

(C) Given the equation $\sqrt{6-5 \cos x+7 \sin ^2 x} = \cos x$. Squaring both sides,we get $6 - 5 \cos x + 7 \sin ^2 x = \cos ^2 x$. Since $\sin ^2 x

Vedclass · Accepted Answer

$	an x+\sec x=1$

Vedclass · Answer

(C) Given the equation $\sqrt{6-5 \cos x+7 \sin ^2 x} = \cos x$. Squaring both sides,we get $6 - 5 \cos x + 7 \sin ^2 x = \cos ^2 x$. Since $\sin ^2 x = 1 - \cos ^2 x$,we have $6 - 5 \cos x + 7(1 - \cos ^2 x) = \cos ^2 x$. $6 - 5 \cos x + 7 - 7 \cos ^2 x = \cos ^2 x$. $13 - 5 \cos x = 8 \cos ^2 x$,which simplifies to $8 \cos ^2 x + 5 \cos x - 13 = 0$. Let $t = \cos x$. Then $8t^2 + 5t - 13 = 0$. Factoring the quadratic,$(8t + 13)(t - 1) = 0$. Thus,$t = 1$ or $t = -13/8$. Since $-1 \le \cos x \le 1$,we must have $\cos x = 1$,which implies $x = 2n\pi$ for any integer $n$. Now check the options for $x = 2n\pi$: For $x = 2n\pi$,$	an x = 0$,$\cot x$ is undefined,$\sec x = 1$,$\operatorname{cosec} x$ is undefined. Re-evaluating the equation: $\sqrt{6-5(1)+7(0)} - 1 = \sqrt{1} - 1 = 0$. This holds. Checking option $A$: $	an x + \cot x = 0 + 	ext{undefined}$. Checking option $B$: $\cot x + \operatorname{cosec} x = 	ext{undefined}$. Checking option $C$: $	an x + \sec x = 0 + 1 = 1$. Thus,the solution satisfies $	an x + \sec x = 1$.

The general solution of the equation $\sqrt{6-5 \cos x+7 \sin ^2 x}-\cos x=0$ also satisfies the equation

Similar Questions

For a $\Delta ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then the sides $a, b, c$ are in:

If $\alpha, \beta, \gamma$ are angles of a triangle,then $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma - 2 \cos \alpha \cos \beta \cos \gamma$ is

In a triangle $ABC$,if $\cos A \cos B + \sin A \sin B \sin C = 1$,then $\sin A + \sin B + \sin C = $

In $\triangle ABC$,if $\cos A \cdot \cos B \cdot \cos C = \frac{1}{5}$,then $\tan A \tan B + \tan B \tan C + \tan C \tan A = $

If in $\triangle ABC$,$a \tan A + b \tan B = (a + b) \tan \left(\frac{A+B}{2}\right)$,then which of the following holds?

Vedclass Test Series

Exam Paper Generator

Online Exam Module