What is the general solution for ((sin(θ))/(2cos(θ)))-1=0 ?

The general solution for ((sin(θ))/(2cos(θ)))-1=0 is θ=1.10714…+pin

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((sin(θ))/(2cos(θ)))-1=0

(\frac{sin ( θ )}{2 cos ( θ )}) - 1 = 0

θ = 1.10714 \dots + πn

Degrees

θ = 63.43494 \dots^{\circ} + 18 0^{\circ} n

Solution steps

(\frac{sin ( θ )}{2 cos ( θ )}) - 1 = 0

Rewrite using trig identities

\frac{sin ( θ )}{2 cos ( θ )} - 1

Use the basic trigonometric identity:

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

= \frac{tan ( θ )}{2} - 1

\frac{tan ( θ )}{2} - 1 = 0

Move

1

to the right side

\frac{tan ( θ )}{2} - 1 = 0

Add

1

to both sides

\frac{tan ( θ )}{2} - 1 + 1 = 0 + 1

Simplify

\frac{tan ( θ )}{2} = 1

\frac{tan ( θ )}{2} = 1

Multiply both sides by

2

\frac{tan ( θ )}{2} = 1

Multiply both sides by

2

\frac{2 tan ( θ )}{2} = 1 \cdot 2

Simplify

tan (θ) = 2

tan (θ) = 2

Apply trig inverse properties

tan (θ) = 2

General solutions for

tan (θ) = 2

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

θ = arctan (2) + πn

θ = arctan (2) + πn

Show solutions in decimal form

θ = 1.10714 \dots + πn

2 \cdot cos^{2} (2 x) + 9 \cdot cos (4 x) = - 2.4214900000000 \dots

tan (- x) = tan (x)

6 sin (θ) = 6 sin (2 θ)

2 sin (1 5^{\circ}) cos (1 5^{\circ}) = tan (x)

6 sin^{2} (x) + 5 cos (x) = 7

What is the general solution for ((sin(θ))/(2cos(θ)))-1=0 ?
The general solution for ((sin(θ))/(2cos(θ)))-1=0 is θ=1.10714…+pin