TOPICS

Exercise - 3.4

Trigonometric Functions

**Question-1 :-** Find the principal and general solutions of tan x = √3.

We know that tan π/3 = √3 and tan(4π/3) = tan(π + π/3) = tan π/3 = √3 Therefore, principal solutions are π/3 and 4π/3. Now, tan x = tan π/3 x = nπ + π/3, where n ∈ Z.

**Question-2 :-** Find the principal and general solutions of sec x = 2.

We know that sec π/3 = 2 and sec(5π/3) = sec(2π - π/3) = sec π/3 =2 Therefore, principal solutions are π/3 and 5π/3. Now, sec x = sec π/3 cos x = cos π/3 [sec x = 1/cos x] x = 2nπ ± π/3, where n ∈ Z.

**Question-3 :-** Find the principal and general solutions of cot x = -√3.

We know that cot π/6 = √3 cot(5π/6) = cot(π - π/6) = -cot π/6 = -√3 cot(11π/6) = cot(2π - π/6) = -cot π/6 = -√3 Therefore, principal solutions are 5π/6 and 11π/6. Now, cot x = cot(5π/6) tan x = tan(5π/6) [cot x = 1/tan x] x = nπ + 5π/6, where n ∈ Z.

**Question-4 :-** Find the principal and general solutions of cosec x = – 2.

We know that cosec π/6 = 2 cosec 7π/6 = cosec(π + π/6) = -cosec π/6 = -2 cosec 11π/6 = cosec(2π - π/6) = -cosec π/6 = -2 Therefore, principal solutions are 7π/6 and 11π/6. Now, cosec x = cosec(7π/6) sin x = sin(7π/6) [cosec x = 1/sin x] x = nπ + (-1)ⁿ 7π/6, where n ∈ Z.

**Question-5 :-** Find the general solutions of cos 4x = cos 2x.

cos 4x = cos 2x cos 4x - cos 2x = 0 -2 sin 6x/2 sin 2x/2 = 0 [cos a - cos b = -2 sin(a + b)/2 . sin(a - b)/2] sin 3x sin x = 0 sin 3x = 0 or sin x = 0 3x = nπ or x = nπ, where n ∈ Z. x = nπ or x = nπ, where n ∈ Z.

**Question-6 :-** Find the general solutions of cos 3x + cos x – cos 2x = 0.

cos 3x + cos x – cos 2x = 0 2 cos 4x/2 cos 2x/2 - cos 2x = 0 [cos a + cos b = 2 cos(a + b)/2 . cos(a - b)/2] 2 cos 2x cos x - cos 2x = 0 cos 2x(2 cos x - 1) = 0 cos 2x = 0 or 2 cos x - 1 = 0 cos 2x = 0 or cos x = 1/2 2x = π/2 or x = π/3 2x = (2n + 1)π/2 or x = 2nπ ± π/3, where n ∈ Z. x = (2n + 1)π/4 or x = 2nπ ± π/3, where n ∈ Z.

**Question-7 :-** Find the general solutions of sin 2x + cosx = 0.

sin 2x + cosx = 0 2 sin x cos x + cos x = 0 cos x(2 sin x + 1) = 0 cos x = 0 or 2 sin x + 1 = 0 cos x = 0 or sin x = -1/2 x = (2n + 1)π/2, where n ∈ Z. Now, sin x = -1/2 sin x = sin(π + π/6) sin x = sin 7π/6 x = 7π/6 x = nπ + (-1)ⁿ 7π/6, where n ∈ Z. Therefore,x = (2n + 1)π/2 or x = nπ + (-1)ⁿ 7π/6, where n ∈ Z.

**Question-8 :-** Find the general solutions of sec² 2x = 1– tan 2x

sec² 2x = 1 – tan 2x 1 + tan² 2x - 1 + tan 2x = 0 tan² 2x + tan 2x = 0 tan 2x(tan 2x + 1) = 0 tan 2x = 0 or tan 2x + 1 = 0 tan 2x = 0 2x = 0 2x = nπ + 0, where n ∈ Z x = nπ/2, where n ∈ Z Now, tan 2x + 1 = 0 tan 2x = -1 tan 2x = -tan π/4 tan 2x = tan(π - π/4) tan 2x = tan 3π/4 2x = 3π/4 2x = nπ + 3π/4, where n ∈ Z x = nπ/2 + 3π/8, where n ∈ Z Therefore, x = nπ/2 or x = nπ/2 + 3π/8, where n ∈ Z

**Question-9 :-** Find the general solutions of sin x + sin 3x + sin 5x = 0

sin x + sin 3x + sin 5x = 0 (sin x + sin 5x) + sin 3x = 0 2 sin 6x/2 . cos (-4x)/2 + sin 3x = 0 [sin a + sin b = 2 sin (a + b)/2 . cos (a - b)/2] 2 sin 3x . cos (-2x) + sin 3x = 0 2 sin 3x . cos 2x + sin 3x = 0 [cos(-x) = cos x] sin 3x(2 cos 2x + 1) = 0 sin 3x = 0 or 2 cos 2x + 1 = 0 sin 3x = 0 3x = 0 3x = nπ + 0, where n ∈ Z x = nπ/3, where n ∈ Z Now, 2 cos 2x + 1 = 0 cos 2x = -1/2 cos 2x = -cos π/3 cos 2x = cos(π - π/3) cos 2x = cos 2π/3 2x = 2π/3 2x = 2nπ ± 2π/3, where n ∈ Z. x = nπ ± π/3, where n ∈ Z. Therefore, x = nπ/3 or x = nπ ± π/3, where n ∈ Z.

CLASSES