Advertisements
Advertisements
प्रश्न
Find the general solution of the following equation:
Advertisements
उत्तर
Ideas required to solve the problem: The general solution of any trigonometric equation is given as:
sin x = sin y, implies x = nπ + (– 1)ny, where n ∈ Z.
cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.
tan x = tan y, implies x = nπ + y, where n ∈ Z.
Given,
tan x + cot 2x = 0
⇒ tan x − cot 2x
We know that: cot θ = tan (π/2 − θ)
∴ `tan x = -tan (pi/2 - 2x)`
⇒ `tan x = tan (2x - pi/2) {∵ - tan θ = tan -θ}`
If tan x = tan y, then x is given by x = nπ + y, where n ∈ Z.
From above expression, on comparison with standard equation we have
y = `(2x - pi/2)`
∴ x = nπ + 2x − `pi/2`
⇒ `x = pi/2 - npi = pi/2(1 - 2n), "where" n ∈ Z`
APPEARS IN
संबंधित प्रश्न
Find the general solution of cosec x = –2
Find the general solution for each of the following equations sec2 2x = 1– tan 2x
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 \[T_n = \sin^n x + \cos^n x\], prove that \[\frac{T_3 - T_5}{T_1} = \frac{T_5 - T_7}{T_3}\]
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 ∆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
If \[\frac{\pi}{2} < x < \frac{3\pi}{2},\text{ then }\sqrt{\frac{1 - \sin x}{1 + \sin x}}\] is equal to
If tan \[x = - \frac{1}{\sqrt{5}}\] and θ lies in the IV quadrant, then the value of cos x is
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
The value of sin25° + sin210° + sin215° + ... + sin285° + sin290° is
If \[cosec x + \cot x = \frac{11}{2}\], then tan x =
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:
Solve the following equation:
\[\sec x\cos5x + 1 = 0, 0 < x < \frac{\pi}{2}\]
Solve the following equation:
\[5 \cos^2 x + 7 \sin^2 x - 6 = 0\]
Solve the following equation:
\[\sin x - 3\sin2x + \sin3x = \cos x - 3\cos2x + \cos3x\]
Solve the following equation:
3tanx + cot x = 5 cosec x
Write the values of x in [0, π] for which \[\sin 2x, \frac{1}{2}\]
and cos 2x are in A.P.
If \[3\tan\left( x - 15^\circ \right) = \tan\left( x + 15^\circ \right)\] \[0 < x < 90^\circ\], find θ.
If \[\cos x + \sqrt{3} \sin x = 2,\text{ then }x =\]
If \[4 \sin^2 x = 1\], then the values of x are
If \[\cot x - \tan x = \sec x\], then, x is equal to
The number of values of x in [0, 2π] that satisfy the equation \[\sin^2 x - \cos x = \frac{1}{4}\]
The equation \[3 \cos x + 4 \sin x = 6\] has .... solution.
Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°
2 cos2x + 1 = – 3 cos x
Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°
cos 2x = 1 − 3 sin x
Solve the following equations:
2 cos2θ + 3 sin θ – 3 = θ
Solve the following equations:
sin θ + cos θ = `sqrt(2)`
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
