- 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
- 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
- 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 +5π/6 , where n €Z
thus, the general solution is x = nπ + 5π/6, where n €Z
- 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
- 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 = nπ/3 or x = nπ, where n €Z
- 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
- 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 7π/6, where n €Z
Therefore, the general solution is (2n+1) π/2 or nπ+(-1)n 7π/6, n €Z
- 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 + 3π/4, where n €Z
x = nπ/2 + 3π/8, where n €Z
Therefore the general solution is nπ/2 or nπ/2 + 3π/8, where n €Z
- 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 ± 2π/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
