What is the general solution for cos(θ)=cos(arctan(3/4)) ?

The general solution for cos(θ)=cos(arctan(3/4)) is θ=0.64350…+2pin,θ=2pi-0.64350…+2pin

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cos(θ)=cos(arctan(3/4))

cos (θ) = cos (arctan (\frac{3}{4}))

θ = 0.64350 \dots + 2 πn, θ = 2 π - 0.64350 \dots + 2 πn

Solution steps

cos (θ) = cos (arctan (\frac{3}{4}))

cos (arctan (\frac{3}{4})) = \frac{4}{5}

cos (arctan (\frac{3}{4}))

Rewrite using trig identities:

cos (arctan (\frac{3}{4})) = \frac{1 + ( \frac{3}{4} ) ^{2}}{1 + ( \frac{3}{4} ) ^{2}}

Use the following identity:

cos (arctan (x)) = \frac{1 + x ^{2}}{1 + x ^{2}}

= \frac{1 + ( \frac{3}{4} ) ^{2}}{1 + ( \frac{3}{4} ) ^{2}}

= \frac{1 + ( \frac{3}{4} ) ^{2}}{1 + ( \frac{3}{4} ) ^{2}}

Simplify

= \frac{4}{5}

cos (θ) = \frac{4}{5}

Apply trig inverse properties

cos (θ) = \frac{4}{5}

General solutions for

cos (θ) = \frac{4}{5}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{4}{5}) + 2 πn, θ = 2 π - arccos (\frac{4}{5}) + 2 πn

θ = arccos (\frac{4}{5}) + 2 πn, θ = 2 π - arccos (\frac{4}{5}) + 2 πn

Show solutions in decimal form

θ = 0.64350 \dots + 2 πn, θ = 2 π - 0.64350 \dots + 2 πn

13 sin (2 x) - 3 = 3

sin (x) = \frac{4}{9}

sin (x) = \frac{4}{6}

- 4 sin (x) = 0

4 sin^{2} (θ) - sin (θ) - 4 = 0

What is the general solution for cos(θ)=cos(arctan(3/4)) ?
The general solution for cos(θ)=cos(arctan(3/4)) is θ=0.64350…+2pin,θ=2pi-0.64350…+2pin