Advertisements
Advertisements
प्रश्न
Solve the following equations:
2 cos2θ + 3 sin θ – 3 = θ
Advertisements
उत्तर
2 cos2θ + 3 sin θ – 3 = θ
2(1 – sin2θ)+ 3 sin θ – 3 = θ
2 – 2 sin2θ + 3 sin θ – 3 = θ
– 2 sin2θ + 3 sin θ – 1 = θ
2 sin2 θ – 3 sin θ + 1 = θ
2 sin2θ – 2 sin θ – sin θ + 1 = θ
2 sin θ (sin θ – 1) – (sin θ – 1) = θ
(2 sin θ – 1)(sin θ – 1) = 0
2 sin θ – 1 = 0 or sin θ – 1 = θ
sin θ = `1/2` or sin θ = 1
To find the general solution of’ sin θ = `1/2`
sin θ = `1/2`
sin θ = `sin pi/6`
The general solution is θ = `"n"pi + (- 1)^"n" pi/6`, n ∈ Z
To find the general solution of sin θ = 1
sin θ = 1
sin θ = `pii/2`
The general solution is θ = `"n"pi + (- 1)^"n" pi/6`, n ∈ Z
∴ The required solutions are
θ = `"n"pi + (- 1)^"n" pi/6`, n ∈ Z
or
θ = `"n"pi + (- 1)^"n" pi/6`, n ∈ Z
APPEARS IN
संबंधित प्रश्न
If \[x = \frac{2 \sin x}{1 + \cos x + \sin x}\], then prove that
If tan x + sec x = \[\sqrt{3}\], 0 < x < π, then x is equal to
If tan A + cot A = 4, then tan4 A + cot4 A is equal to
The value of \[\cos1^\circ \cos2^\circ \cos3^\circ . . . \cos179^\circ\] is
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:
\[\sqrt{3} \cos x + \sin x = 1\]
If cos x = k has exactly one solution in [0, 2π], then write the values(s) of k.
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
Find the principal solution and general solution of the following:
cot θ = `sqrt(3)`
Solve the following equations:
sin 5x − sin x = cos 3
Solve the following equations:
sin θ + sin 3θ + sin 5θ = 0
Solve the following equations:
sin 2θ – cos 2θ – sin θ + cos θ = θ
Solve the following equations:
sin θ + cos θ = `sqrt(2)`
Solve 2 tan2x + sec2x = 2 for 0 ≤ x ≤ 2π.
