# Answer to Question #9944 in Trigonometry for Eliana Hernandez

Question #9944

solve each on the interval (0, 2pi)

1. tan(x)sec(x)=3tan(x)

2. tan^2(x)-3tan(x)-2=0

1. tan(x)sec(x)=3tan(x)

2. tan^2(x)-3tan(x)-2=0

Expert's answer

1. tan(x)sec(x)=3tan(x)

tan(x)/cos(x)=3tan(x)

a)tan(x)=0 <=> sin(x)=0 => answer on the required interval is x=pi

b) tan(x)<>0

cos(x)=1/3

x==-arccos(1/3)+2pik where k-ineteger

final answer is combined from answers a),b)

2. tan^2(x)-3tan(x)-2=0

Let's denote& tan(x)=t

t^2-3t-2=0

t=(3+sqrt(17))/2,t=(3-sqrt(17))/2

a)tan(x)=(3+sqrt(17))/2

x=arctan((3+sqrt(17))/2)+pi k, k-integer

b) tan(x)=(3-sqrt(17))/2

x=arctan((3-sqrt(17))/2)+pi n , n-integerfinal answer is combined from answers a),b)

tan(x)/cos(x)=3tan(x)

a)tan(x)=0 <=> sin(x)=0 => answer on the required interval is x=pi

b) tan(x)<>0

cos(x)=1/3

x==-arccos(1/3)+2pik where k-ineteger

final answer is combined from answers a),b)

2. tan^2(x)-3tan(x)-2=0

Let's denote& tan(x)=t

t^2-3t-2=0

t=(3+sqrt(17))/2,t=(3-sqrt(17))/2

a)tan(x)=(3+sqrt(17))/2

x=arctan((3+sqrt(17))/2)+pi k, k-integer

b) tan(x)=(3-sqrt(17))/2

x=arctan((3-sqrt(17))/2)+pi n , n-integerfinal answer is combined from answers a),b)

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