Advertisements
Advertisements
प्रश्न
Solve the following equations:
2cos 2x – 7 cos x + 3 = 0
Advertisements
उत्तर
2 cos2x – 7 cos x + 3 = 0
2 cos2x – 6 cos x – cos x + 3 = 0
2 cos x (cos x – 3) – 1(cos x – 3) = 0
(2 cos x – 1)(cos x – 3) = 0
2 cos x – 1 = 0 or cos x – 3 = 0
cos x = `1/2` or cos x = 3|
Since – 1 ≤ cos x ≤ 1, we have
cos x = 3 is not possible.
∴ cos x = `1/2`
cos x = `cos pi/3`
The general solution is x = `2"n"pi +- pi/3`, n ∈ Z
APPEARS IN
संबंधित प्रश्न
If \[x = \frac{2 \sin x}{1 + \cos x + \sin x}\], then prove that
If \[\sin x = \frac{a^2 - b^2}{a^2 + b^2}\], then the values of tan x, sec x and cosec x
If \[\tan x = \frac{a}{b},\] show that
If \[cosec x - \sin x = a^3 , \sec x - \cos x = b^3\], then prove that \[a^2 b^2 \left( a^2 + b^2 \right) = 1\]
Prove the:
\[ \sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}} = - \frac{2}{\cos x},\text{ where }\frac{\pi}{2} < x < \pi\]
Prove that
Find x from the following equations:
\[cosec\left( \frac{\pi}{2} + \theta \right) + x \cos \theta \cot\left( \frac{\pi}{2} + \theta \right) = \sin\left( \frac{\pi}{2} + \theta \right)\]
Find the general solution of the following equation:
Find the general solution of the following equation:
Find the general solution of the following equation:
Find the general solution of the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
\[\sqrt{3} \cos x + \sin x = 1\]
Write the values of x in [0, π] for which \[\sin 2x, \frac{1}{2}\]
and cos 2x are in A.P.
The general solution of the equation \[7 \cos^2 x + 3 \sin^2 x = 4\] is
The solution of the equation \[\cos^2 x + \sin x + 1 = 0\] lies in the interval
Choose the correct alternative:
If f(θ) = |sin θ| + |cos θ| , θ ∈ R, then f(θ) is in the interval
Find the general solution of the equation 5cos2θ + 7sin2θ – 6 = 0
Number of solutions of the equation tan x + sec x = 2 cosx lying in the interval [0, 2π] is ______.
