Find the number of solutions of the equation | cot x | = cot x + 1 sin x , when x ∈ [ 0 , 2 π ]

If cot x > 0 , then we get 1 sin x = 0 which has no solution. ⋯ 1 Now if cot x < 0 then - cot x = cot x + 1 sin x ⇒ 2 cos x + 1 sin x = 0 ⇒ cos x = - 1 2 ⇒ x = 2 n π ± 2 π 3 , n ∈ I and 0 ≤ x ≤ 2 π Therefore x = 2 π 3 , 4 π 3 . But at x = 4 π 3 , cot x > 0 and according to 1 it will not be the solution. Hence, it has only one solution i.e. x = 2 π 3 .

Find the number of solutions of the equation | cot x | = cot x + 1 sin x , when x ∈ [ 0 , 2 π ]

If cot x > 0 , then we get 1 sin x = 0 which has no solution. ⋯ 1 Now if cot x < 0 then - cot x = cot x + 1 sin x ⇒ 2 cos x + 1 sin x = 0 ⇒ cos x = - 1 2 ⇒ x = 2 n π ± 2 π 3 , n ∈ I and 0 ≤ x ≤ 2 π Therefore x = 2 π 3 , 4 π 3 . But at x = 4 π 3 , cot x > 0 and according to 1 it will not be the solution. Hence, it has only one solution i.e. x = 2 π 3 .

Find the number of solutions of the equations: sin 3 x cos x + sin 2 x · cos 2 x + sin x · cos 3 x = 1 , when x ∈ [ 0 , 2 π ] .

From the question sin 3 x cos x + sin 2 x · cos 2 x + sin x · cos 3 x = 1 After taking sin x cos x common, we get ⇒ sin x cos x sin 2 x + sin x cos x + cos 2 x = 1 ⇒ sin 2 x 2 1 + sin 2 x 2 = 1 ⇒ sin 2 x 2 + sin 2 x = 4 ⇒ sin 2 2 x + 2 sin 2 x - 4 = 0 ⇒ sin 2 x = - 2 ± 4 + 16 2 = - 1 ± 5 These values are not in the range of sine function. As - 1 ≤ sin 2 x ≤ 1 . Hence, the given equation has no solution.

Find the number of solutions of the equation: x + 2 tan x = π 2 , when x ∈ [ 0 , 2 π ]

Given equation is x + 2 tan x = π 2 tan x = π 4 - x 2 If we plot the graph of the curve y = tan x and y = π 4 - x 2 in the interval [ 0 , 2 π ] intersect at three points. Hence, number of solutions is three.

Find the set of all x in ( - π , π ) satisfying | 4 sin x - 1 | < 5 .

Given equation | 4 sin x - 1 | < 5 ( now use x < a ⇒ - a < x < a ) ⇒ - 5 < 4 sin x - 1 < 5 ⇒ - 5 + 1 4 < sin x < 5 + 1 4 Principal argument for which sin x = - 5 + 1 4 is - π 10 . Principal argument for which sin x = 5 + 1 4 is 3 π 10 . Hence, it is clear from the graph that it has 3 intervals for the solution. x ∈ - π , - 9 π 10 ∪ - π 10 , 3 π 10 ∪ 7 π 10 , π

If x ∈ [ 0 , π ] and 81 sin 2 x + 81 cos 2 x = 30 , then find x .

Let 81 sin 2 x = y Then 81 cos 2 x = 81 1 - sin 2 x = 81 . 81 - sin 2 x = 81 · y - 1 Now the given question can be written as y 2 - 30 y + 81 = 0 Using factorization, we get ⇒ y = 3 or y = 27 ⇒ 81 sin 2 x = 3 or 27 ⇒ 3 4 sin 2 x = 3 or 3 3 ⇒ sin 2 x = 1 4 or 3 4 when x ∈ 0 , π ⇒ x = π 6 , π 3 , 5 π 6 , 2 π 3

If x ∈ [ 0 , π ] and 3 sin 2 x + 2 cos 2 x + 3 1 - sin 2 x + 2 sin 2 x = 28 , then find x .

We have 3 sin 2 x + 2 cos 2 x + 3 1 - sin 2 x + 2 sin 2 x = 28 or 3 sin 2 x + 2 cos 2 x + 3 3 - sin 2 x + 2 cos 2 x = 28 Let 3 sin 2 x + 2 cos 2 x = y Now given question can be written as y + 27 y = 28 ⇒ y 2 - 28 y + 27 = 0 y = 27 , 1 ⇒ 3 sin 2 x + 2 cos 2 x = 3 3 or ⇒ 3 sin 2 x + 2 cos 2 x = 3 0 ⇒ sin 2 x + 2 cos 2 x = 3 or sin 2 x + 2 cos 2 x = 0 ⇒ sin 2 x + cos 2 x = 2 or 2 cos x ( sin x + cos x ) = 0 ⇒ sin 2 x = cos 2 x = 1 or ( cos x = 0 or tan x = - 1 ) Which is not possible that sin 2 x = cos 2 x = 1 , for any value of x and so y ≠ 27 . ⇒ cos x = 0 or tan x = - 1 ⇒ x = π 2 , 3 π 4

Solve the equation: cos 3 x · cos 3 x + sin 3 x · sin 3 x = 0

We have cos 3 x · cos 3 x + sin 3 x · sin 3 x = 0 Using cos 3 x = 4 cos 3 x - 3 cos x , sin 3 x = - 4 sin 3 x + 3 sin x. 1 4 cos 2 3 x + 3 cos x cos 3 x + 3 sin x sin 3 x - sin 2 3 x = 0 ⇒ 1 4 [ 3 cos 2 x + cos 6 x ] = 0 ⇒ cos 3 2 x = 0 (Put 2 x = t and use cos 3 t + 3 cos t = 4 cos 3 t ) ⇒ 2 x = 2 n + 1 π 2 ⇒ x = 2 n + 1 π 4