"Trigonometric EquationPrincipal Solutions1. sin x=√32A. 5π62. tan x=−1√3B. 5π123. sec 2x=−2√3C. 7π124. cos 3x=(−12)D. π3E. 2π9F. 4π9G. 11π6H. 2π3 "
1 - d, 2 - a, 3 - b, 4 - e
1 - d, 2 - g, 3 - b, 4 - f
1 - h, 2 - a, 3 - c, 4 - j
1 - h, 2 - g, 3 - c, 4 - e
We have learnt that the values of sin x &cos x repeat after an interval of 2π and the values of tan x repeat after an interval of pi. The solutions of trigonometric equations for which x lies between 0 and 2π are called principal solutions.
So,
1.sinx=√32
We know sin x is positive in i & ii quadrant in the interval of 0 to 2 π.
in i quadrant, sinπ3=(√3)2
in ii quadrant sin(π−π3)=sin2π3=√32
Therefore, principal solution of equation sinx=√32areπ3,2π3
2.tanx=−1√3
tan x is negative in ii & iv quadrant in the interval of 0 to 2π.
We know that tanπ6=1√3. tan(π−π6)=−tanπ6=−1√3
So,π−π6=5π6and
in iv quadrant (2π−π6)=−tanπ6=−1√3
(2π−π6)=11π6
Therefore, Principal solutions are 5π6and11π6
3.sec2xc=−2√3
1cos2x=−2√3
cos2x=−√32
Cosπ is negative in ii & iii quadrant in the interval of 0 to 2π.
We know, cos π6 = √32
in ii quadrant cos(π−π6)=−cos(π6)=√32
π−π6=5π6
2x=5π6
x=5π12
In iii quadrant (π−π6)=−cosπ6=−√32
2x=π+π6=7π6
x=7π12
Therefore, principal solutions of equations sec2x=−2√3 are 5π12 & 7π2
cos3x=−12
cosθ is negative in ii & iii quadrant
cosπ3=12
In ii quadrant cos(π−π3)=−cosπ3=−12
3x=π−π3=2π3
x=2π9
In iii quadrant cos(π−π3)=−cosπ3=−12
3x=π−π3=4π3
x=4π9
Therefore, principal solutions of equation cos3x=−12 are 2π9 & 4π9