Advertisements
Advertisements
प्रश्न
Solve the following equation:
Advertisements
उत्तर
\[2 \cos^2 x - 5 \cos x + 2 = 0\]
\[ \Rightarrow 2 \cos^2 x - 4 \cos x - \cos x + 2 = 0\]
\[ \Rightarrow 2 \cos x ( \cos x - 2) - 1 ( \cos x - 2) = 0\]
\[ \Rightarrow (\cos x - 2) ( 2 \cos\theta - 1) = 0\]
\[\therefore 2 \cos x - 1 = 0 \]
\[ \Rightarrow \cos x = \frac{1}{2} \]
\[ \Rightarrow \cos x = \cos \frac{\pi}{3} \]
\[ \Rightarrow x = 2n\pi \pm \frac{\pi}{3}, n \in Z\]
APPEARS IN
संबंधित प्रश्न
If \[\sin x = \frac{a^2 - b^2}{a^2 + b^2}\], then the values of tan x, sec x and cosec x
If \[cosec x - \sin x = a^3 , \sec x - \cos x = b^3\], then prove that \[a^2 b^2 \left( a^2 + b^2 \right) = 1\]
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 \[ab + a - b + 1 = 0\]
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 \[2 T_6 - 3 T_4 + 1 = 0\]
If \[T_n = \sin^n x + \cos^n x\], prove that \[6 T_{10} - 15 T_8 + 10 T_6 - 1 = 0\]
Prove that: \[\tan\frac{11\pi}{3} - 2\sin\frac{4\pi}{6} - \frac{3}{4} {cosec}^2 \frac{\pi}{4} + 4 \cos^2 \frac{17\pi}{6} = \frac{3 - 4\sqrt{3}}{2}\]
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:
\[\sin\frac{13\pi}{3}\sin\frac{8\pi}{3} + \cos\frac{2\pi}{3}\sin\frac{5\pi}{6} = \frac{1}{2}\]
Prove that:
Prove that:
If \[\frac{3\pi}{4} < \alpha < \pi, \text{ then }\sqrt{2\cot \alpha + \frac{1}{\sin^2 \alpha}}\] is equal to
sin6 A + cos6 A + 3 sin2 A cos2 A =
If x sin 45° cos2 60° = \[\frac{\tan^2 60^\circ cosec30^\circ}{\sec45^\circ \cot^{2^\circ} 30^\circ}\], then x =
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:
Solve the following equation:
Solve the following equation:
4sinx cosx + 2 sin x + 2 cosx + 1 = 0
Solve the following equation:
3tanx + cot x = 5 cosec x
Solve the following equation:
3 – 2 cos x – 4 sin x – cos 2x + sin 2x = 0
Solve the following equation:
\[2^{\sin^2 x} + 2^{\cos^2 x} = 2\sqrt{2}\]
If secx cos5x + 1 = 0, where \[0 < x \leq \frac{\pi}{2}\], find the value of x.
Write the number of solutions of the equation
\[4 \sin x - 3 \cos x = 7\]
If \[3\tan\left( x - 15^\circ \right) = \tan\left( x + 15^\circ \right)\] \[0 < x < 90^\circ\], find θ.
If a is any real number, the number of roots of \[\cot x - \tan x = a\] in the first quadrant is (are).
If \[4 \sin^2 x = 1\], then the values of x are
If \[\sqrt{3} \cos x + \sin x = \sqrt{2}\] , then general value of x is
General solution of \[\tan 5 x = \cot 2 x\] is
Find the principal solution and general solution of the following:
sin θ = `-1/sqrt(2)`
Solve the following equations:
sin θ + sin 3θ + sin 5θ = 0
Choose the correct alternative:
If tan 40° = λ, then `(tan 140^circ - tan 130^circ)/(1 + tan 140^circ * tan 130^circ)` =
Choose the correct alternative:
If tan α and tan β are the roots of x2 + ax + b = 0 then `(sin(alpha + beta))/(sin alpha sin beta)` is equal to
Choose the correct alternative:
If f(θ) = |sin θ| + |cos θ| , θ ∈ R, then f(θ) is in the interval
