Advertisements
Advertisements
प्रश्न
Solve the following equations:
cos θ + cos 3θ = 2 cos 2θ
उत्तर
cos 3θ + cos θ = 2 cos 2θ
`2 cos ((3theta + theta)/2) * cos ((3theta - theta)/2)` = 2 cos 2θ
`2cos ((4theta)/2) * cos ((2theta)/2)` = 2 cos 2θ
2 cos 2θ . cos θ = 2 cos 2θ
cos 2θ . cos θ – cos 2θ = θ
cos 2θ (cos θ – 1) = θ
cos 2θ = θ or cos θ – 1 = θ
cos 2θ = θ or cos θ = 1
To find the general solution of cos 2θ = θ
The general solution is
2θ = `(2"n" + 1) pi/2`, n ∈ Z
θ = `(2"n" + 1) pi/4`, n ∈ Z
To find the general solution of cos θ = 1
cos θ = 1
cos θ = cos 0
The general solution is θ = 2nπ , n ∈ Z
∴ The required solutions are
θ = `(2"n" + 1) pi/4`, n ∈ Z
or
θ = 2nπ, n ∈ Z
APPEARS IN
संबंधित प्रश्न
Find the general solution of the equation cos 3x + cos x – cos 2x = 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 ∆ABC, prove that:
In a ∆ABC, 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 x from the following equations:
\[x \cot\left( \frac{\pi}{2} + \theta \right) + \tan\left( \frac{\pi}{2} + \theta \right)\sin \theta + cosec\left( \frac{\pi}{2} + \theta \right) = 0\]
Prove that:
\[\tan 4\pi - \cos\frac{3\pi}{2} - \sin\frac{5\pi}{6}\cos\frac{2\pi}{3} = \frac{1}{4}\]
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:
4sinx cosx + 2 sin x + 2 cosx + 1 = 0
Solve the following equation:
3tanx + cot x = 5 cosec x
The solution of the equation \[\cos^2 x + \sin x + 1 = 0\] lies in the interval
Find the principal solution and general solution of the following:
cot θ = `sqrt(3)`
Choose the correct alternative:
If tan 40° = λ, then `(tan 140^circ - tan 130^circ)/(1 + tan 140^circ * tan 130^circ)` =
Solve 2 tan2x + sec2x = 2 for 0 ≤ x ≤ 2π.
