What is the general solution for tan(2θ)=-2.5 ?

The general solution for tan(2θ)=-2.5 is θ=-(1.19028…)/2+(pin)/2

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tan(2θ)=-2.5

tan (2 θ) = - 2.5

θ = - \frac{1.19028 \dots}{2} + \frac{πn}{2}

Degrees

θ = - 34.09929 \dots^{\circ} + 9 0^{\circ} n

Solution steps

tan (2 θ) = - 2.5

Apply trig inverse properties

tan (2 θ) = - 2.5

General solutions for

tan (2 θ) = - 2.5

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

2 θ = arctan (- 2.5) + πn

2 θ = arctan (- 2.5) + πn

Solve

2 θ = arctan (- 2.5) + πn : θ = - \frac{arctan ( \frac{5}{2} )}{2} + \frac{πn}{2}

2 θ = arctan (- 2.5) + πn

Simplify

arctan (- 2.5) + πn : - arctan (\frac{5}{2}) + πn

arctan (- 2.5) + πn

arctan (- 2.5) = - arctan (\frac{5}{2})

arctan (- 2.5)

= arctan (- \frac{5}{2})

Use the following property:

arctan (- x) = - arctan (x)

arctan (- \frac{5}{2}) = - arctan (\frac{5}{2})

= - arctan (\frac{5}{2})

= - arctan (\frac{5}{2}) + πn

2 θ = - arctan (\frac{5}{2}) + πn

Divide both sides by

2

2 θ = - arctan (\frac{5}{2}) + πn

Divide both sides by

2

\frac{2 θ}{2} = - \frac{arctan ( \frac{5}{2} )}{2} + \frac{πn}{2}

Simplify

θ = - \frac{arctan ( \frac{5}{2} )}{2} + \frac{πn}{2}

θ = - \frac{arctan ( \frac{5}{2} )}{2} + \frac{πn}{2}

θ = - \frac{arctan ( \frac{5}{2} )}{2} + \frac{πn}{2}

Show solutions in decimal form

θ = - \frac{1.19028 \dots}{2} + \frac{πn}{2}

1 + 4 cos (θ) = 3 sin (θ), 0 \leq θ \leq 2 π

\frac{150}{sin ( x )} = \frac{140.78}{sin ( 11 5 ^{\circ} )}

tan^{2} (x) + cot^{2} (x) = 2

tan^{2} (x) + cot^{2} (x) = 1

4 sin (θ) = - 4 cos (θ)

What is the general solution for tan(2θ)=-2.5 ?
The general solution for tan(2θ)=-2.5 is θ=-(1.19028…)/2+(pin)/2