Advertisements
Advertisements
प्रश्न
Solve the following equations:
sin θ + cos θ = `sqrt(2)`
Advertisements
उत्तर
Divide each term by `sqrt(2)`
`1/sqrt(2) sin theta + 1/sqrt(2) cos theta = sqrt(2)/sqrt(2)`
`sin pi/4 sin theta + cos pi/4 cos theta` = 1
`cos theta * cos pi/4 + sin theta * sin pi/4` = 1
`cos (theta - pi/4)` = 1
`cos (theta - pi/4)` = cos θ
The general solution is
`theta - pi/4` = 2nπ, n ∈ Z
θ = `2"n"pi + pi/4`, n ∈ Z
θ = `(8"n"pi + pi)/4`, n ∈ Z
θ = `(8"n" + 1) pi/4`, n ∈ Z
APPEARS IN
संबंधित प्रश्न
Find the principal and general solutions of the equation `cot x = -sqrt3`
If \[x = \frac{2 \sin x}{1 + \cos x + \sin x}\], then prove that
If \[\tan x = \frac{a}{b},\] show 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:
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:
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 + sec x = \[\sqrt{3}\], 0 < x < π, then x is equal to
sin6 A + cos6 A + 3 sin2 A cos2 A =
Find the general solution of the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
\[2 \sin^2 x = 3\cos x, 0 \leq x \leq 2\pi\]
The general solution of the equation \[7 \cos^2 x + 3 \sin^2 x = 4\] is
If \[4 \sin^2 x = 1\], then the values of x are
Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°
sin4x = sin2x
Solve the following equations:
2 cos2θ + 3 sin θ – 3 = θ
Solve the following equations:
sin 2θ – cos 2θ – sin θ + cos θ = θ
Find the general solution of the equation 5cos2θ + 7sin2θ – 6 = 0
