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

The general solution for cos(2x)=tan(2x) is x=(0.66623…)/2+pin,x= pi/2-(0.66623…)/2+pin

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cos(2x)=tan(2x)

cos (2 x) = tan (2 x)

x = \frac{0.66623 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.66623 \dots}{2} + πn

Solution steps

cos (2 x) = tan (2 x)

Subtract

tan (2 x)

from both sides

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

Express with sin, cos

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

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

cos^{2} (2 x) - sin (2 x) = 0

Rewrite using trig identities

1 - sin (2 x) - sin^{2} (2 x) = 0

Solve by substitution

sin (2 x) = - \frac{1 + 5}{2}, sin (2 x) = \frac{5 - 1}{2}

sin (2 x) = - \frac{1 + 5}{2} :

No Solution

sin (2 x) = \frac{5 - 1}{2} : x = \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn

Combine all the solutions

x = \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn

Show solutions in decimal form

x = \frac{0.66623 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.66623 \dots}{2} + πn

sin (\frac{θ}{2}) - 1 = 0

2 sin (4 x) + 3 = 0

7 sec (x) + 7 tan (x) = 7

sin (2 x) = - 2 sin (x)

10 sec^{2} (x) + 5 tan^{2} (x) - 15 = 0

What is the general solution for cos(2x)=tan(2x) ?
The general solution for cos(2x)=tan(2x) is x=(0.66623…)/2+pin,x= pi/2-(0.66623…)/2+pin