What is the general solution for 2tan(x)+sec(x)=0,0<= x<2pi ?

The general solution for 2tan(x)+sec(x)=0,0<= x<2pi is x=(7pi)/6 ,x=(11pi)/6

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2tan(x)+sec(x)=0,0<= x<2pi

2 tan (x) + sec (x) = 0, 0 \leq x < 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

Solution steps

2 tan (x) + sec (x) = 0, 0 \leq x < 2 π

Express with sin, cos

\frac{1 + 2 sin ( x )}{cos ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

1 + 2 sin (x) = 0

Move

1

to the right side

2 sin (x) = - 1

Divide both sides by

2

sin (x) = - \frac{1}{2}

General solutions for

sin (x) = - \frac{1}{2}

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

1 - sin (x) = 3 cos (x)

cos (x) - sec (x) = 0

2 + 7 cos (x) = 4 sin^{2} (x)

4 arccos (x) = π

2 sin^{2} (x) + 2 = 7 cos (x)

What is the general solution for 2tan(x)+sec(x)=0,0<= x<2pi ?
The general solution for 2tan(x)+sec(x)=0,0<= x<2pi is x=(7pi)/6 ,x=(11pi)/6