What is the general solution for 2cos(2x)+sin(2x)=0 ?

The general solution for 2cos(2x)+sin(2x)=0 is x=-(1.10714…)/2+(pin)/2

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

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

x = - \frac{1.10714 \dots}{2} + \frac{πn}{2}

Degrees

x = - 31.71747 \dots^{\circ} + 9 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Divide both sides by

cos (2 x), cos (2 x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

2 + tan (2 x) = 0

2 + tan (2 x) = 0

Move

2

to the right side

2 + tan (2 x) = 0

Subtract

2

from both sides

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

Simplify

tan (2 x) = - 2

tan (2 x) = - 2

Apply trig inverse properties

tan (2 x) = - 2

General solutions for

tan (2 x) = - 2

tan (x) = - a \Rightarrow x = arctan (- a) + πn

2 x = arctan (- 2) + πn

2 x = arctan (- 2) + πn

Solve

2 x = arctan (- 2) + πn : x = - \frac{arctan ( 2 )}{2} + \frac{πn}{2}

2 x = arctan (- 2) + πn

Simplify

arctan (- 2) + πn : - arctan (2) + πn

arctan (- 2) + πn

Use the following property:

arctan (- x) = - arctan (x)

arctan (- 2) = - arctan (2)

= - arctan (2) + πn

2 x = - arctan (2) + πn

Divide both sides by

2

2 x = - arctan (2) + πn

Divide both sides by

2

\frac{2 x}{2} = - \frac{arctan ( 2 )}{2} + \frac{πn}{2}

Simplify

x = - \frac{arctan ( 2 )}{2} + \frac{πn}{2}

x = - \frac{arctan ( 2 )}{2} + \frac{πn}{2}

x = - \frac{arctan ( 2 )}{2} + \frac{πn}{2}

Show solutions in decimal form

x = - \frac{1.10714 \dots}{2} + \frac{πn}{2}

csc^{2} (θ) - cot (θ) = 1

sin (x) = \frac{2}{13}

- sec^{2} (x) = 0

2 cos (x) + 4 cos (4 x) = 0

tan (θ) = \frac{24}{7}

What is the general solution for 2cos(2x)+sin(2x)=0 ?
The general solution for 2cos(2x)+sin(2x)=0 is x=-(1.10714…)/2+(pin)/2