Question

# "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 "

A

1 - d, 2 - a, 3 - b, 4 - e

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B

1 - d, 2 - g, 3 - b, 4 - f

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C

1 - h, 2 - a, 3 - c, 4 - j

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D

1 - h, 2 - g, 3 - c, 4 - e

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Solution

## The correct options are A 1 - d, 2 - a, 3 - b, 4 - e B 1 - d, 2 - g, 3 - b, 4 - f C 1 - h, 2 - a, 3 - c, 4 - j D 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

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