General Solution of cos theta = cos alpha
Trending Questions
Q. The general solution of sin2x+sin4x+sin6x=0 is/are
- nπ±π6, n∈Z
- nπ4, n∈Z
- 2nπ±2π3, n∈Z
- nπ±π3, n∈Z
Q. If 4sin2θ+2(√3+1)cosθ=4+√3, then the general solution is
- nπ±π3, n∈Z
- nπ±π6, n∈Z
- 2nπ±π3, n∈Z
- 2nπ±π4, n∈Z
Q. If sin3θ=sinθ, how many solutions exist such that −2π<θ<2π ?
- 8
- 9
- 11
- 7
Q.
___
Total number of solutions of equation sin x tan 4x = cos x belonging to (0, π) are:
Q. The number of solution(s) of sin3x−sinx=4cos2x−2 for x∈[0, 2π] is
- 1
- 2
- More than 2 but finite
- infinitely many
Q. The sum of all the solutions of the equation cosθcos(π3+θ)cos(π3−θ)=14, θ∈[0, 6π] is
- 15π
- 30π
- 100π3
- None of these
Q. The general solution of sin2x+sin4x+sin6x=0 is/are
- nπ±π6, n∈Z
- nπ4, n∈Z
- 2nπ±2π3, n∈Z
- nπ±π3, n∈Z
Q.
Find the general solution of (2 sin x - cos x) (1 + cos x) = sin2 x
nπ + (−1)n π6
(2n + 1)π
2nπ
nπ + (−1)n π3
Q.
Evaluate: .
Q.
How do you use trigonometric identities to transform one side of the equation into the other ?
Q.
___
Find number of value of xϵ[0, π] satisfying the relation cos 3x + sin 2x - sin 4x = 0
Q. The general solution of the equation cosθ=cosα, is given by θ=nπ±α, n∈Z.
- True
- False
Q. If sin3θ=sinθ, how many solutions exist such that −2π<θ<2π ?
- 8
- 9
- 11
- 7
Q. If 3sinx+4cosax=7 has at least one solution and a=pmqn+1; where m, n∈Z, then the value of p+q is
Q.
Solve 2cos2θ+cosθ−1=0
2nπ ∀ n∈N
2nπ∓π3 ∀ n∈N
2nπ∓π6 ∀ n∈N
(2n+1)π6 ∀ n∈N
Q. cos(α−β)=1 and cos(α+β)=1e, where α, β∈[−π, π]. Number of pairs of α, β which satisfy both the equation is
- \N
- 1
- 2
- 4
Q. If the equation 2cosx+cos2λx=3 has only one solution, then λ is
- 1
- a rational number
- an irrational number
- None of these
Q. If 3cosx≠2sinx, then the general solution of sin2x−cos2x=2−sin2x is
- nπ+(−1)nπ2, n∈Z
- nπ2, n∈Z
- (4n±1)π2, n∈Z
- (2n−1)π, n∈Z
Q. The general value of θ in the equation cosθ=1√2, tanθ=−1 is
- 2nπ±π6, n∈I
- 2nπ±7π4, n∈I
- nπ+(−1)n.π3, n∈I
- nπ+(−1)n.π4, n∈I
Q. General values of x for which sin2x+cosx=0 is/are :
- 2nπ3+π6
- 2nπ3−π6
- 2nπ+π2
- 2nπ−π2
Q. If 3cosx≠2sinx, then the general solution of sin2x−cos2x=2−sin2x is
- nπ+(−1)nπ2, n∈Z
- nπ2, n∈Z
- (4n±1)π2, n∈Z
- (2n−1)π, n∈Z