Advertisements
Advertisements
प्रश्न
Solve the following equation:
\[2 \sin^2 x = 3\cos x, 0 \leq x \leq 2\pi\]
Advertisements
उत्तर
\[2 \sin^2 x = 3\cos x\]
\[ \Rightarrow 2\left( 1 - \cos^2 x \right) = 3\cos x\]
\[ \Rightarrow 2 \cos^2 x + 3\cos x - 2 = 0\]
\[ \Rightarrow \left( 2\cos x - 1 \right)\left( \cos x + 2 \right) = 0\]
\[\Rightarrow \cos x = \frac{1}{2} \text{ or }\cos x = - 2\]
But,
\[\cos x = - 2\] is not possible.
`therefore cosx=1/2=cos pi/3`
Putting n = 0 and n = 1, we get
APPEARS IN
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
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 \[6 T_{10} - 15 T_8 + 10 T_6 - 1 = 0\]
Prove that
Prove that:
\[\sec\left( \frac{3\pi}{2} - x \right)\sec\left( x - \frac{5\pi}{2} \right) + \tan\left( \frac{5\pi}{2} + x \right)\tan\left( x - \frac{3\pi}{2} \right) = - 1 .\]
Find x from the following equations:
\[x \cot\left( \frac{\pi}{2} + \theta \right) + \tan\left( \frac{\pi}{2} + \theta \right)\sin \theta + cosec\left( \frac{\pi}{2} + \theta \right) = 0\]
If \[cosec x - \cot x = \frac{1}{2}, 0 < x < \frac{\pi}{2},\]
If \[cosec x + \cot x = \frac{11}{2}\], then tan x =
If tan θ + sec θ =ex, then cos θ equals
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:
Find the general solution of the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
\[\sin x - 3\sin2x + \sin3x = \cos x - 3\cos2x + \cos3x\]
Solve the following equation:
3tanx + cot x = 5 cosec x
If secx cos5x + 1 = 0, where \[0 < x \leq \frac{\pi}{2}\], find the value of x.
Write the values of x in [0, π] for which \[\sin 2x, \frac{1}{2}\]
and cos 2x are in A.P.
If \[2 \sin^2 x = 3\cos x\]. where \[0 \leq x \leq 2\pi\], then find the value of x.
If \[\cos x + \sqrt{3} \sin x = 2,\text{ then }x =\]
If \[\tan px - \tan qx = 0\], then the values of θ form a series in
The number of solution in [0, π/2] of the equation \[\cos 3x \tan 5x = \sin 7x\] is
General solution of \[\tan 5 x = \cot 2 x\] is
The solution of the equation \[\cos^2 x + \sin x + 1 = 0\] lies in the interval
If \[\cos x = - \frac{1}{2}\] and 0 < x < 2\pi, then the solutions are
Find the principal solution and general solution of the following:
sin θ = `-1/sqrt(2)`
Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°
sin4x = sin2x
Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°
2 sin2x + 1 = 3 sin x
Solve the following equations:
sin 5x − sin x = cos 3
Solve the following equations:
sin θ + sin 3θ + sin 5θ = 0
Solve `sqrt(3)` cos θ + sin θ = `sqrt(2)`
If a cosθ + b sinθ = m and a sinθ - b cosθ = n, then show that a2 + b2 = m2 + n2
