# Trigonometry Functions – Class XI – Exercise 3.4

1. Find the principal and general solutions of the equation tanx = √3

Solution:

tanx = √3

We know that,

tan π/3 = √3 and tan4π/3 = tan(π + π/3) = tan π/3 = √3

therefore, the principal solutions are x = π/3 and 4π/3

Now, tanx = tan π/3

x =n π+ π/3, where n € Z

Therefore, the general solution is x = n π + π/3, where n € Z

1. Find the principal and general solutions of the equation secx = 2

Solution:

secx = 2

We know that secπ/3 = 2 and sec 5π/3 = sec(2π – π/3) =secπ/3 = 2

therefore, the principal solutions are x = π/3 and 5π/3

Now, sec = secπ/3

cosx = cosπ/3

x = 2nπ± π/3, where n  € Z

Thus, the general so;lution is x = 2nπ± π/3, where n  € Z

1. Find the principal and general solutions of the equation cotx = -√3

Solution:

cot x = -√3

We know that cot π/6 =√3

cot(π – π/6) = -cotπ/6 = -√3 and cot(2π – π/6) = -cotπ/6 = -√3

i.e., cot 5π/6 = -√3

and cot 11π/6 = -√3

Thus the principal solutions are x = 5π/6 and 11π/6

Now, cotx = cot 5π/6

tanx = tan 5π/6               [cot x = 1/tanx}

x = nπ +/6 , where n €Z

thus, the general solution is x = nπ + /6, where n €Z

1. find the general solution of cosecx = -2

solution:

We know that cosec π/6 = 2

cosec(π + π/6) = -cosecπ/6 = -2

and cosec(2π-π/6) = -cosecπ/6 = -2

i.e., cosec7π/6 = -2 and cosec11π/6 = -2

Therefore the principal solutions are x = 7π/6 and 11π/6

Now, cosecx = cosec7π/6

sinx = sin7π/6                         [cosecx = 1/sinx]

x = nπ + (-1)n  7π/6, where n €Z

Therefore, the general solution is x = nπ + (-1)n  7π/6, where n €Z

1. Find the general solution of the equation cos4x = cos 2x

Solution:

cos4x = cos2x

cos4x – cos2x = 0

-2sin(4x+2x/2)sin(4x-2x/2)=0

[cosA-cosB = -2sin(A+B/2)sin(A-B/2)]

sin3x.sinx = 0

sin3x = 0  or  sinx =0

3x = nπ or x = nπ, where n €Z

x = /3 or x = nπ, where n €Z

1. Find the general solution of the equation cos3x+cosx-cos2x = 0

Solution:

cos3x + cosx – cos2x = 0

2cos(3x+x/2)cos(3x-x/2) – cos2x = 0

[cosA +cosB = 2cos(A+B/2)cos(A-B/2)]

2cos2x.cosx – cos2x =0

cos2x(2cosx – 1) = 0

cos 2x =0        or    2cosx-1 = 0

cos2x = 0        or      cosx = 1/2

2x = (2n+1) π/2        or      cosx = cosπ/3, where n €Z

x = (2n+1) π/3                 or           x = 2nπ±π/3, where n €Z

1. Find the general solution of the equation sin2x + cosx = 0

Solution:

sin 2x + cos x = 0

2sinxcosx + cosx = 0

cosx(2sinx + 1) = 0

cosx = 0             or       2sinx+1=0

Now, cosx = 0 ; cosx = (2n+1) π/2 , where n €Z

2sinx + 1 = 0

sinx = –1/2 = -sinπ/6 = sin(π+π/6) = sin 7π/6

x = nπ + (-1)n  /6, where n €Z

Therefore, the general solution is (2n+1) π/2 or nπ+(-1)n  /6, n €Z

1. Find the general solution of the equation sec2 2x = 1 – tan2x

Solution:

sec22x = 1 – tan2x

1+tan22x = 1 – tan2x

tan22x + tan2x = 0

tan 2x(tan2x+1) = 0

tan 2x = 0             or               tan2x+1 = 0

Now, tan 2x = 0

tan 2x = tan0

2x = nπ+0, where n €Z

tan2x + 1 = 0

tan 22x = -1 = -tanπ/4 = tan(π-π/4) = tan 3π/4

2x = nπ + /4, where n €Z

x = nπ/2 + /8, where n €Z

Therefore the general solution is /2 or nπ/2 + /8, where n €Z

1. Find the general solution of the equation sinx + sin3x+ sin5x = 0

Solution:

sinx + sin3x+ sin5x = 0

(sinx + sin5x) + sin3x = 0

[2sin(x+5x/2)cos(x-5x/2)]+ sin3x = 0

sinA+sinB = 2sin(A+B/2)cos(A-B/2)]

2sin3xcos(-2x)+sin3x = 0

2sin3x cos2x + sin3x = 0

sin3x(2cos2x+1)=0

sin33x = 0 or 2cos2x+1 =0

Now, sin3x = 0 ; 3x = n π , where n €Z

i.e., x = n π/3, where n €Z

2cos2x + 1 = 0

cos2x = –1/2 = -cos π/3 = cos(π- π/3)

cos2x = cos2 π/3

2x = 2n π ±/3, where n €Z

x = n π ± π/3, where n €Z

Therefore the general solution is nπ or n π ± π/3, where n €Z