Advertisements
Advertisements
Question
Find the principal solution and general solution of the following:
sin θ = `-1/sqrt(2)`
Advertisements
Solution
sin θ = `-1/sqrt(2)`
We know that principal of sin θ lies in `[ - pi/2, pi/2]`
sin θ = `- 1/sqrt(2 < 0`
∴ The principal value of sin θ lies in the IV quadrant.
sin θ = `- 1/sqrt(2)`
= `- sin(pi/4)`
sin θ = `sin (- pi/4)`
Hence θ = `- pi/4` is the principal solution.
The general solution is
θ = nπ + (– 1)n . `( pi/4)`, n ∈ Z
θ = `"n"pi + (- 1)^("n" + 1) * pi/4`, n ∈ Z
APPEARS IN
RELATED QUESTIONS
If \[\sin x + \cos x = m\], then prove that \[\sin^6 x + \cos^6 x = \frac{4 - 3 \left( m^2 - 1 \right)^2}{4}\], where \[m^2 \leq 2\]
If \[a = \sec x - \tan x \text{ and }b = cosec x + \cot x\], then shown that \[ab + a - b + 1 = 0\]
Prove that
Prove that:
\[\tan 4\pi - \cos\frac{3\pi}{2} - \sin\frac{5\pi}{6}\cos\frac{2\pi}{3} = \frac{1}{4}\]
Which of the following is correct?
Find the general solution of the following equation:
Solve the following equation:
\[\sin^2 x - \cos x = \frac{1}{4}\]
Solve the following equation:
Solve the following equation:
Solve the following equation:
\[\sqrt{3} \cos x + \sin x = 1\]
Solve the following equation:
\[2 \sin^2 x = 3\cos x, 0 \leq x \leq 2\pi\]
Solve the following equation:
\[\sin x - 3\sin2x + \sin3x = \cos x - 3\cos2x + \cos3x\]
Write the number of points of intersection of the curves
A solution of the equation \[\cos^2 x + \sin x + 1 = 0\], lies in the interval
The smallest positive angle which satisfies the equation
If \[\cot x - \tan x = \sec x\], then, x is equal to
A value of x satisfying \[\cos x + \sqrt{3} \sin x = 2\] is
If \[\sqrt{3} \cos x + \sin x = \sqrt{2}\] , then general value of x is
Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°
2 sin2x + 1 = 3 sin x
