What is the general solution for tan(θ)= 2/(3+2cos(45)) ?

The general solution for tan(θ)= 2/(3+2cos(45)) is θ=0.42541…+180n

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tan(θ)= 2/(3+2cos(45))

tan (θ) = \frac{2}{3 + 2 cos ( 4 5 ^{\circ} )}

θ = 0.42541 \dots + 18 0^{\circ} n

Radians

θ = 0.42541 \dots + πn

Solution steps

tan (θ) = \frac{2}{3 + 2 cos ( 4 5 ^{\circ} )}

cos (4 5^{\circ}) = \frac{2}{2}

cos (4 5^{\circ})

Use the following trivial identity:

cos (4 5^{\circ}) = \frac{2}{2}

cos (4 5^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

= \frac{2}{2}

tan (θ) = \frac{2}{3 + 2 \cdot \frac{2}{2}}

Simplify

\frac{2}{3 + 2 \cdot \frac{2}{2}} : \frac{2 ( 3 - 2 )}{7}

\frac{2}{3 + 2 \cdot \frac{2}{2}}

2 \cdot \frac{2}{2} = 2

2 \cdot \frac{2}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 \cdot 2}{2}

Cancel the common factor:

2

= 2

= \frac{2}{3 + 2}

Rationalize

\frac{2}{3 + 2} : \frac{2 ( 3 - 2 )}{7}

\frac{2}{3 + 2}

Multiply by the conjugate

\frac{3 - 2}{3 - 2}

= \frac{2 ( 3 - 2 )}{( 3 + 2 ) ( 3 - 2 )}

(3 + 2) (3 - 2) = 7

(3 + 2) (3 - 2)

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 3, b = 2

= 3^{2} - (2)^{2}

Simplify

3^{2} - (2)^{2} : 7

3^{2} - (2)^{2}

3^{2} = 9

3^{2}

3^{2} = 9

= 9

(2)^{2} = 2

(2)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 2

= 9 - 2

Subtract the numbers:

9 - 2 = 7

= 7

= 7

= \frac{2 ( 3 - 2 )}{7}

= \frac{2 ( 3 - 2 )}{7}

tan (θ) = \frac{2 ( 3 - 2 )}{7}

Apply trig inverse properties

tan (θ) = \frac{2 ( 3 - 2 )}{7}

General solutions for

tan (θ) = \frac{2 ( 3 - 2 )}{7}

tan (x) = a \Rightarrow x = arctan (a) + 18 0^{\circ} n

θ = arctan (\frac{2 ( 3 - 2 )}{7}) + 18 0^{\circ} n

θ = arctan (\frac{2 ( 3 - 2 )}{7}) + 18 0^{\circ} n

Show solutions in decimal form

θ = 0.42541 \dots + 18 0^{\circ} n

cos (2 θ) + 6 cos^{2} (θ) = 5

sin (4 x) = - \frac{2}{2}

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

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

sin (x) = - \frac{12}{13}

What is the general solution for tan(θ)= 2/(3+2cos(45)) ?
The general solution for tan(θ)= 2/(3+2cos(45)) is θ=0.42541…+180n