Advertisements
Advertisements
प्रश्न
The equation \[3 \cos x + 4 \sin x = 6\] has .... solution.
पर्याय
finite
infinite
one
no
Advertisements
उत्तर
no
Given equation:
\[3 \cos x + 4 \sin x = 6\] ...(i)
Thus, the equation is of the form
\[a \cos x + b \sin x = c\], where
\[a = 3, b = 4\] and c = 6.
Let: \[a = 3 = r \cos \alpha\] and \[b = 4 = r \sin \alpha\]
Now,
\[\tan \alpha = \frac{b}{a} = \frac{4}{3}\]
\[ \Rightarrow \alpha = \tan^{- 1} \left( \frac{4}{3} \right)\]
Also,
\[r = \sqrt{a^2 + b^2} = \sqrt{9 + 16} = \sqrt{25} = 5\]
On putting
\[a = 3 = r \cos \alpha\] and \[b = 4 = r \sin \alpha\] in equation (i), we get:
\[r \cos\alpha \cos\theta + \sin\alpha \sin\theta = 6\]
\[ \Rightarrow r \cos (\theta - \alpha ) = 6\]
\[ \Rightarrow 5 \cos (\theta - \alpha) = 6\]
\[ \Rightarrow \cos (\theta - \alpha) = \frac{6}{5}\]
From here, we cannot find the value of \[\theta\]
APPEARS IN
संबंधित प्रश्न
Find the general solution of the equation cos 4 x = cos 2 x
Find the general solution of the equation sin x + sin 3x + sin 5x = 0
If \[x = \frac{2 \sin x}{1 + \cos x + \sin x}\], then prove that
If \[\sin x = \frac{a^2 - b^2}{a^2 + b^2}\], then the values of tan x, sec x and cosec x
If \[\tan x = \frac{a}{b},\] show that
If \[T_n = \sin^n x + \cos^n x\], prove that \[2 T_6 - 3 T_4 + 1 = 0\]
Prove that
Find x from the following equations:
\[cosec\left( \frac{\pi}{2} + \theta \right) + x \cos \theta \cot\left( \frac{\pi}{2} + \theta \right) = \sin\left( \frac{\pi}{2} + \theta \right)\]
If tan x = \[x - \frac{1}{4x}\], then sec x − tan x is equal to
If sec \[x = x + \frac{1}{4x}\], then sec x + tan x =
If x = r sin θ cos ϕ, y = r sin θ sin ϕ and z = r cos θ, then x2 + y2 + z2 is independent of
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
If tan θ + sec θ =ex, then cos θ equals
The value of \[\tan1^\circ \tan2^\circ \tan3^\circ . . . \tan89^\circ\] is
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:
\[\cot x + \tan x = 2\]
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:
cosx + sin x = cos 2x + sin 2x
Solve the following equation:
3 – 2 cos x – 4 sin x – cos 2x + sin 2x = 0
Solve the following equation:
3sin2x – 5 sin x cos x + 8 cos2 x = 2
If cos x = k has exactly one solution in [0, 2π], then write the values(s) of k.
The general value of x satisfying the equation
\[\sqrt{3} \sin x + \cos x = \sqrt{3}\]
If \[\cot x - \tan x = \sec x\], then, x is equal to
If \[\sqrt{3} \cos x + \sin x = \sqrt{2}\] , then general value of x is
Find the principal solution and general solution of the following:
cot θ = `sqrt(3)`
Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°
2 cos2x + 1 = – 3 cos x
If 2sin2θ = 3cosθ, where 0 ≤ θ ≤ 2π, then find the value of θ.
Find the general solution of the equation sinx – 3sin2x + sin3x = cosx – 3cos2x + cos3x
In a triangle ABC with ∠C = 90° the equation whose roots are tan A and tan B is ______.
