मराठी

Solve the Following Equation: 4 Sin 2 X − 8 Cos X + 1 = 0 - Mathematics

Advertisements
Advertisements

प्रश्न

Solve the following equation:

\[4 \sin^2 x - 8 \cos x + 1 = 0\]
बेरीज

उत्तर

\[4 \sin^2 x - 8\cos x + 1 = 0\]
\[\Rightarrow 4 - 4 \cos^2 x - 8 \cos x + 1 = 0\]
\[ \Rightarrow 4 \cos^2 x + 8 \cos x - 5 = 0\]
\[ \Rightarrow 4 \cos^2 x + 10 \cos x - 2 \cos x - 5 = 0\]
\[ \Rightarrow 2 \cos x (2 \cos x + 5 ) - 1 (2 \cos x + 5) = 0\]
\[ \Rightarrow (2 \cos x - 1) (2 \cos x + 5) = 0\]
\[\Rightarrow (2 \cos x - 1) = 0\] or
\[(2 \cos x + 5) = 0\]
Now,
\[2 \cos x + 5 = 0 \Rightarrow \cos x = - \frac{5}{2}\]  (It is not possible.)

\[\therefore 2 \cos x - 1 = 0 \]

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

\[ \Rightarrow \cos x = \cos \frac{\pi}{3} \]

\[ \Rightarrow x = 2n\pi \pm \frac{\pi}{3}, n \in Z\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Trigonometric equations - Exercise 11.1 [पृष्ठ २२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 11 Trigonometric equations
Exercise 11.1 | Q 3.4 | पृष्ठ २२

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find the principal and general solutions of the equation `tan x = sqrt3`


Find the general solution of the equation cos 4 x = cos 2 x


If \[\cot x \left( 1 + \sin x \right) = 4 m \text{ and }\cot x \left( 1 - \sin x \right) = 4 n,\] \[\left( m^2 + n^2 \right)^2 = mn\]


If \[a = \sec x - \tan x \text{ and }b = cosec x + \cot x\], then shown that  \[ab + a - b + 1 = 0\]


In a ∆A, B, C, D be the angles of a cyclic quadrilateral, taken in order, prove that cos(180° − A) + cos (180° + B) + cos (180° + C) − sin (90° + D) = 0


Find x from the following equations:
\[cosec\left( \frac{\pi}{2} + \theta \right) + x \cos \theta \cot\left( \frac{\pi}{2} + \theta \right) = \sin\left( \frac{\pi}{2} + \theta \right)\]


Prove that:

\[\tan\frac{5\pi}{4}\cot\frac{9\pi}{4} + \tan\frac{17\pi}{4}\cot\frac{15\pi}{4} = 0\]

 


If x = r sin θ cos ϕ, y = r sin θ sin ϕ and r cos θ, then x2 + y2 + z2 is independent of


If \[cosec x - \cot x = \frac{1}{2}, 0 < x < \frac{\pi}{2},\]

 

If x is an acute angle and \[\tan x = \frac{1}{\sqrt{7}}\], then the value of \[\frac{{cosec}^2 x - \sec^2 x}{{cosec}^2 x + \sec^2 x}\] is


If A lies in second quadrant 3tan A + 4 = 0, then the value of 2cot A − 5cosA + sin A is equal to


The value of \[\cos1^\circ \cos2^\circ \cos3^\circ . . . \cos179^\circ\] is

 

Which of the following is correct?


Find the general solution of the following equation:

\[\tan 2x \tan x = 1\]

Find the general solution of the following equation:

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

Find the general solution of the following equation:

\[\sin x = \tan x\]

Solve the following equation:

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

Solve the following equation:

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

Solve the following equation:
\[\cot x + \tan x = 2\]

 


Solve the following equation:
3 – 2 cos x – 4 sin x – cos 2x + sin 2x = 0


Write the number of solutions of the equation
\[4 \sin x - 3 \cos x = 7\]


If \[2 \sin^2 x = 3\cos x\]. where \[0 \leq x \leq 2\pi\], then find the value of x.


If \[\tan px - \tan qx = 0\], then the values of θ form a series in

 


The general solution of the equation \[7 \cos^2 x + 3 \sin^2 x = 4\] is


The number of solution in [0, π/2] of the equation \[\cos 3x \tan 5x = \sin 7x\] is 


The general value of x satisfying the equation
\[\sqrt{3} \sin x + \cos x = \sqrt{3}\]


If \[e^{\sin x} - e^{- \sin x} - 4 = 0\], then x =


General solution of \[\tan 5 x = \cot 2 x\] is


The number of values of x in the interval [0, 5 π] satisfying the equation \[3 \sin^2 x - 7 \sin x + 2 = 0\] is


Find the principal solution and general solution of the following:
cot θ = `sqrt(3)`


Solve the following equations:
2 cos2θ + 3 sin θ – 3 = θ


Solve the following equations:
`sin theta + sqrt(3) cos theta` = 1


Solve the following equations:
cot θ + cosec θ = `sqrt(3)`


Solve the following equations:
`tan theta + tan (theta + pi/3) + tan (theta + (2pi)/3) = sqrt(3)`


Find the general solution of the equation 5cos2θ + 7sin2θ – 6 = 0


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×