English

Solve the Following Equation: 5 Cos 2 X + 7 Sin 2 X − 6 = 0

Advertisements
Advertisements

Question

Solve the following equation:
\[5 \cos^2 x + 7 \sin^2 x - 6 = 0\]

Sum
Advertisements

Solution

\[5 \cos^2 x + 7 \sin^2 x - 6 = 0\]
\[ \Rightarrow 5 \cos^2 x + 7\left( 1 - \cos^2 x \right) - 6 = 0\]
\[ \Rightarrow - 2 \cos^2 x + 1 = 0\]
\[ \Rightarrow \cos^2 x = \frac{1}{2} = \cos^2 \frac{\pi}{4}\]
\[ \Rightarrow x = n\pi \pm \frac{\pi}{4}, n \in Z \left( \cos^2 x = \cos^2 \alpha \Rightarrow x = n\pi \pm \alpha, n \in Z \right)\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 11: Trigonometric equations - Exercise 11.1 [Page 22]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 11 Trigonometric equations
Exercise 11.1 | Q 7.4 | Page 22

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

If \[\sin x = \frac{a^2 - b^2}{a^2 + b^2}\], then the values of tan x, sec x and cosec x


If \[\tan x = \frac{b}{a}\] , then find the values of \[\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}\].


Prove the:
\[ \sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}} = - \frac{2}{\cos x},\text{ where }\frac{\pi}{2} < x < \pi\]


If \[T_n = \sin^n x + \cos^n x\], prove that \[\frac{T_3 - T_5}{T_1} = \frac{T_5 - T_7}{T_3}\]

 


If \[T_n = \sin^n x + \cos^n x\], prove that  \[2 T_6 - 3 T_4 + 1 = 0\]


Prove that:  tan 225° cot 405° + tan 765° cot 675° = 0


Prove that: tan (−225°) cot (−405°) −tan (−765°) cot (675°) = 0


Prove that cos 570° sin 510° + sin (−330°) cos (−390°) = 0

Prove that:

\[3\sin\frac{\pi}{6}\sec\frac{\pi}{3} - 4\sin\frac{5\pi}{6}\cot\frac{\pi}{4} = 1\]

 


Prove that

\[\frac{cosec(90^\circ + x) + \cot(450^\circ + x)}{cosec(90^\circ - x) + \tan(180^\circ - x)} + \frac{\tan(180^\circ + x) + \sec(180^\circ - x)}{\tan(360^\circ + x) - \sec( - x)} = 2\]

 


Prove that:
\[\sin^2 \frac{\pi}{18} + \sin^2 \frac{\pi}{9} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{4\pi}{9} = 2\]

 

In a ∆ABC, prove that:
cos (A + B) + cos C = 0


In a ∆ABC, prove that:

\[\tan\frac{A + B}{2} = \cot\frac{C}{2}\]

Prove that:
\[\sin \frac{13\pi}{3}\sin\frac{2\pi}{3} + \cos\frac{4\pi}{3}\sin\frac{13\pi}{6} = \frac{1}{2}\]


If sec \[x = x + \frac{1}{4x}\], then sec x + tan x = 

 

If tan x + sec x = \[\sqrt{3}\], 0 < x < π, then x is equal to


If \[\frac{3\pi}{4} < \alpha < \pi, \text{ then }\sqrt{2\cot \alpha + \frac{1}{\sin^2 \alpha}}\] is equal to


The value of sin25° + sin210° + sin215° + ... + sin285° + sin290° is


The value of \[\tan1^\circ \tan2^\circ \tan3^\circ . . . \tan89^\circ\] is

 

Find the general solution of the following equation:

\[\cos 3x = \frac{1}{2}\]

Find the general solution of the following equation:

\[\sin 3x + \cos 2x = 0\]

Solve the following equation:

\[2 \sin^2 x + \sqrt{3} \cos x + 1 = 0\]

Solve the following equation:

\[\sin x + \sin 2x + \sin 3x + \sin 4x = 0\]

Solve the following equation:

\[\tan x + \tan 2x + \tan 3x = 0\]

Solve the following equation:

\[\sqrt{3} \cos x + \sin x = 1\]


Write the general solutions of tan2 2x = 1.

 

Write the set of values of a for which the equation

\[\sqrt{3} \sin x - \cos x = a\] has no solution.

Write the number of values of x in [0, 2π] that satisfy the equation \[\sin x - \cos x = \frac{1}{4}\].


If \[\cos x + \sqrt{3} \sin x = 2,\text{ then }x =\]

 


The smallest positive angle which satisfies the equation ​

\[2 \sin^2 x + \sqrt{3} \cos x + 1 = 0\] is

The number of values of ​x in [0, 2π] that satisfy the equation \[\sin^2 x - \cos x = \frac{1}{4}\]


Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°

2 cos2x + 1 = – 3 cos x


Solve the following equations:
sin θ + cos θ = `sqrt(2)`


Choose the correct alternative:
`(cos 6x + 6 cos 4x + 15cos x + 10)/(cos 5x + 5cs 3x + 10 cos x)` is equal to


If sin θ and cos θ are the roots of the equation ax2 – bx + c = 0, then a, b and c satisfy the relation ______.


In a triangle ABC with ∠C = 90° the equation whose roots are tan A and tan B is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×