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

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

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

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

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

Degrees

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

Solution steps

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

Rewrite using trig identities

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

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

tan (2 x) + 2 = 0

tan (2 x) + 2 = 0

Move

2

to the right side

tan (2 x) + 2 = 0

Subtract

2

from both sides

tan (2 x) + 2 - 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}

cos (x) = \frac{4}{12}

\frac{sin ( x )}{2} = \frac{cos ( x )}{3}

4 sin^{2} (θ) - 2 = 0, \frac{π}{2} < θ < \frac{3 π}{2}

9 sin (x) + 6 cos (x) = 10

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

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