What is the general solution for tan(2x)=2cos(x) ?

The general solution for tan(2x)=2cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin

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tan(2x)=2cos(x)

tan (2 x) = 2 cos (x)

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Solution steps

tan (2 x) = 2 cos (x)

Subtract

2 cos (x)

from both sides

tan (2 x) - 2 cos (x) = 0

Express with sin, cos

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

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

sin (2 x) - 2 cos (2 x) cos (x) = 0

Rewrite using trig identities

- 2 cos (2 x) cos (x) + 2 cos (x) sin (x) = 0

Factor

- 2 cos (2 x) cos (x) + 2 cos (x) sin (x) : 2 cos (x) (- cos (2 x) + sin (x))

2 cos (x) (- cos (2 x) + sin (x)) = 0

Solving each part separately

cos (x) = 0 or - cos (2 x) + sin (x) = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

- cos (2 x) + sin (x) = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

3 sec^{2} (θ) - 2 tan^{2} (θ) - 4 = 0

cos^{2} (x) + 5 sin^{2} (x) = 2

(sin (x))^{(s i n (x))} = \frac{1}{2}

(d^{2} - 4) = cos^{2} (x)

cos (3 θ) = 4 cos (3 θ) - 3 cos (θ)

What is the general solution for tan(2x)=2cos(x) ?
The general solution for tan(2x)=2cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin