Advertisements
Advertisements
प्रश्न
Find the general solutions of the following equation:
sin θ - cos θ = 1
Advertisements
उत्तर
sin θ − cos θ = 1
∴ cos θ − sin θ = −1
∴ (1) cos θ − (1) sin θ = −1
`sqrt((1)^2 + (1)^2) = sqrt(1 + 1) = sqrt2`
dividing b. s. by `sqrt2`
∴ `1/sqrt2 costheta - 1/sqrt2 sintheta = -1/sqrt2`
∴ `cos pi/4 costheta - sin pi/4 sintheta = - cos pi/4`
∴ `cos"A" cos"B" - sin"A" sin"B" = cos"(A + B)"`
∴ `cos(theta + pi/4) = cos (pi - pi/4) ...(∵ - costheta = cos(pi - theta))`
∴ `cos (theta + pi/4) = cos (3pi)/4`
cos θ = cos α ⇒ θ = 2n π ± α, n ∈ 2
∴ `theta + pi/4 = 2npi +- (3pi)/4, n ∈ 2`
∴ `theta = 2npi +- (3pi)/4 - pi/4, n ∈ 2`
∴ `theta = 2n pi + (3pi)/4 - pi/4 or theta = 2npi - (3pi)/4 - pi/4, n ∈ 2`
∴ `theta = 2npi + (2pi)/4 or theta = 2npi - (4pi)/4 - pi/4, n ∈ 2`
∴ `theta = 2npi + pi/2 or theta = 2npi - pi n ∈ 2`
∴ These are required general solutions.
APPEARS IN
संबंधित प्रश्न
Find the principal solution of the following equation :
cot θ = `sqrt(3)`
Find the principal solution of the following equation:
cot θ = 0
Find the principal solution of the following equation:
sin θ = `-1/2`
Find the principal solution of the following equation:
`sqrt(3)` cosecθ + 2 = 0
Find the general solution of the following equation:
sinθ = `1/2`.
Find the general solution of the following equation:
tan θ = - 1
Find the general solution of the following equation:
sin 2θ = `1/2`
Find the general solution of the following equation:
tan `(2θ)/(3) = sqrt3`
Find the general solution of the following equation:
4sin2θ = 1.
Find the general solution of the following equation:
cos 4θ = cos 2θ
Find the general solution of the following equation:
tan3θ = 3 tanθ.
Find the general solution of the following equation:
cosθ + sinθ = 1.
State whether the following equation has a solution or not?
2sinθ = 3
In ΔABC, if ∠A = 45°, ∠B = 60° then find the ratio of its sides.
In Δ ABC, prove that a3 sin(B – C) + b3sin(C – A) + c3sin(A – B) = 0
Select the correct option from the given alternatives:
The principal solutions of equation cot θ = `sqrt3` are ______.
Select the correct option from the given alternatives:
The general solution of sec x = `sqrt(2)` is ______.
If in a triangle, the angles are in A.P. and b: c = `sqrt3: sqrt2`, then A is equal to
If `"sin"^-1 4/5 + "cos"^-1 12/13 = "sin"^-1 alpha`, then α = ______.
Select the correct option from the given alternatives:
`2 "tan"^-1 (1/3) + "tan"^-1 (1/7) =` _____
In Δ ABC, prove that `cos(("A" - "B")/2) = (("a" + "b")/"c")sin "C"/2` .
In ΔABC, prove that `("a - b")^2 cos^2 "C"/2 + ("a + b")^2 sin^2 "C"/2 = "c"^2`
In Δ ABC, if cos A = sin B - cos C then show that it is a right-angled triangle.
State whether the following equation has a solution or not?
3 sin θ = 5
If 2 tan-1(cos x) = tan-1(2 cosec x), then find the value of x.
Show that `cot^-1 1/3 - tan^-1 1/3 = cot^-1 3/4`.
Show that `tan^-1 1/2 = 1/3 tan^-1 11/2`
If x, y, z are positive, then prove that
`tan^-1 (("x - y")/(1 + "xy")) + tan^-1 (("y - z")/(1 + "yz")) + tan^-1 (("z - x")/(1 + "zx")) = 0`
Find the principal solutions of sin x − 1 = 0
Find the principal solutions of sin x = `-1/2`
If cos–1x + cos–1y – cos–1z = 0, then show that x2 + y2 + z2 – 2xyz = 1
If f(x) = sin-1`(sqrt((1 - x)/2))`, then f'(x) = ?
If y = sin-1 `[(sqrt(1 + x) + sqrt(1 - x))/2]`, then `"dy"/"dx"` = ?
If 2 cos2 θ + 3 cos θ = 2, then permissible value of cos θ is ________.
Which of the following equations has no solution?
If `(tan 3 theta - 1)/(tan 3 theta + 1) = sqrt3`, then the general value of θ is ______.
The general solution of 4sin2 x = 1 is ______.
The number of solutions of sin x + sin 3x + sin 5x = 0 in the interval `[pi/2, (3pi)/2]` is ______.
If y = `(2sinα)/(1 + cosα + sinα)`, then value of `(1 - cos α + sin α)/(1 + sin α)` is ______.
Which of the following equation has no solution?
The number of principal solutions of tan 2θ = 1 is ______.
Principal solutions at the equation sin 2x + cos 2x = 0, where π < x < 2 π are ______.
If a = sin θ + cos θ, b = sin3 θ + cos3 θ, then ______.
If `sin^-1 4/5 + cos^-1 12/13` = sin–1α, then find the value of α.
