General Solution of Trigonometric Equation
Trending Questions
Prove the following:
cos(π4−x)cos(π4−y)- sin(π4−x)sin(π4−y) = sin (x+y)
- x=nπ, n∈Z
- x=2nπ±π2, n∈Z
- x=(2n+1)π2, n∈Z
- x=ϕ
If tan1∘tan2∘.............tan89∘=x2−8, then the value of x can be
±4
±2
±3
±1
The number of values of y in [−2π, 2π] satisfying the equation |sin 2x| = |cos 2x| = |sin y| is
3
4
1
2
If for a △ABC, cotA. cotB. cotC>0 then the triangle is
All of these
Right angled
Acute angled
Obtuse angled
(where n∈Z)
- nπ+π4
- nπ+π6
- nπ−π6
- 2nπ+π2
- 1
- 2
- 3
- no solution
If log|sinx||cosx|+log|cosx||sinx|=2 then |tanx|
0
2
3
1
- nπ2, n∈Z
- nπ, n∈Z
- (3n±1)π3, n∈Z
- (3n±1)π9, n∈Z
- nπ+π4, n∈Z
- ϕ
- nπ±π4, n∈Z
- 2nπ+π4, n∈Z
The complete solution set of the inequality [cot−1x]2−6[cot−1x]+9≤0, where [.] denotes greatest integer function is
- (−∞, cot2)
- (−∞, cot3)
- (cot2, cot3)
- (cot2, ∞)
- {nπ+π4}
- {nπ+π6}
- {nπ+π3}
- ϕ
- (0, π4)∪(π2, 3π4)
- (0, π2)∪(π2, 3π4)
- (π4, π2)∪(π2, 3π4)
- (0, π4)∪(3π4, π)
- π6
- 7π6
- 11π6
- 7π6, 11π6
- [π2, 5π6]∪[π, 3π2]
- [π2, 2π3]∪[π, 3π2]
- [2π3, 5π6]∪[π, 3π2]
- [π, 3π2]
1 - Q, 2 - P, 3 - S, 4 - R
1 - Q, 2 - P, 3 - Q, 4 - R
1 - R, 2 - R, 3 - Q, 4 - R
1 - R, 2 - P, 3 - S, 4 - P
1tana+1tanb+1tanc+1tand is
- a∈R
- a∈[−e4, −14]
- a∈[−1−e24e, ∞]
- no real value of a exists
(where n∈Z)
- (2n+1)π2
- (4n−1)π2
- (2n−1)π2
- (4n+1)π2
- (−π, −4π5)∪(4π5, π)
- (−4π5, −π5)∪(π5, 4π5)
- (−3π5, −π5)∪(π5, 3π5)
- (−π, −π5)∪(π5, 3π5)
- nπ3+π9, n∈Z
- nπ+π9, n∈Z
- nπ−π9, n∈Z
- nπ2+π3, n∈Z
- π2
- 3π2
- 2
- 1
- (2n+1)π12, n∈Z
- (2n+1)π10, n∈Z
- (2n+1)π2, n∈Z
- (2n+1)π4, n∈Z
- 5
- 7
- 10
- 6
Point D, E are taken on the side BC of the triangle ABC, such that BD = DE = EC. If ∠BAD=x, ∠DAE=y, ∠EAC=z, then the value of sin(x+y)sin(y+z)sinx sinz
2
4
3
1
- 3π4
- 5π4
- 11π4
- 13π4
The complete solution set of the inequality [cot−1x]2−6[cot−1x]+9≤0, where [.] denotes greatest integer function is
(−∞, cot2)
(−∞, cot3)
(cot2, cot3)
(cot2, ∞)
- nπ±π20, n∈Z
- nπ5+π20, n∈Z
- nπ5±π4, n∈Z
- nπ5±π20, n∈Z