Advertisements
Advertisements
Question
Solve the following equations:
sin 5x − sin x = cos 3
Advertisements
Solution
`2cos ((5x + x)/2) * sin((5x - x)/2)` = cos 3x
`2 cos (6x/2) * sin (4x/2)` = cos 3x
2 cos 3 x . sin 2x = cos 3x
2 cos 3x . sin 2x – cos 3x = 0
cos 3x (2 sin 2x – 1) = 0
cos 3x = 0 or 2 sin 2x – 1 = 0
cos 3x = 0 or sin 2x = `1/2`
To find the general solution of cos 3x = 0
The general solution of cos 3x = 0 is
3x = `(2"n" + 1)^(pi/2)`, n ∈ Z
x = `(2"n" + 1)^(pi/6)`, n ∈ Z
To find the general solution of sin 2x = `1/2`
sin 2x = `1/2`
sin 2x = `sin (pi/6)`
The general solution is
2x = `"n"pi + (- 1)^"n" pi/6`, n ∈ Z
x = `("n"pi)/2 + (- 1)^"n" pi/12`, n ∈ Z
∴ The required solutions are
x = `(2"n" + 1) pi/6`, n ∈ Z
x = `("n"pi)/2 + (- 1)^"n" pi/12`, n ∈ Z
APPEARS IN
RELATED QUESTIONS
Find the general solution of the equation sin x + sin 3x + sin 5x = 0
If \[\tan x = \frac{a}{b},\] show that
Prove that
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 ∆ABC, prove that:
If tan x = \[x - \frac{1}{4x}\], then sec x − tan x is equal to
If \[cosec x - \cot x = \frac{1}{2}, 0 < x < \frac{\pi}{2},\]
The value of sin25° + sin210° + sin215° + ... + sin285° + sin290° is
Solve the following equation:
\[\sin^2 x - \cos x = \frac{1}{4}\]
Solve the following equation:
Solve the following equation:
\[2 \sin^2 x = 3\cos x, 0 \leq x \leq 2\pi\]
A solution of the equation \[\cos^2 x + \sin x + 1 = 0\], lies in the interval
The number of solution in [0, π/2] of the equation \[\cos 3x \tan 5x = \sin 7x\] is
The number of values of x in the interval [0, 5 π] satisfying the equation \[3 \sin^2 x - 7 \sin x + 2 = 0\] is
Solve the following equations:
`tan theta + tan (theta + pi/3) + tan (theta + (2pi)/3) = sqrt(3)`
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
