What is the general solution for tan(2θ)+6tan(θ)+8=0 ?

The general solution for tan(2θ)+6tan(θ)+8=0 is θ=-0.64552…+pin,θ=0.81999…+pin,θ=-1.02643…+pin

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tan(2θ)+6tan(θ)+8=0

tan (2 θ) + 6 tan (θ) + 8 = 0

θ = - 0.64552 \dots + πn, θ = 0.81999 \dots + πn, θ = - 1.02643 \dots + πn

Solution steps

tan (2 θ) + 6 tan (θ) + 8 = 0

Rewrite using trig identities

8 + \frac{- 6 tan ^{3} ( θ ) + 8 tan ( θ )}{1 - tan ^{2} ( θ )} = 0

Solve by substitution

tan (θ) \approx - 0.75317 \dots, tan (θ) \approx 1.07169 \dots, tan (θ) \approx - 1.65186 \dots

tan (θ) = - 0.75317 \dots : θ = arctan (- 0.75317 \dots) + πn

tan (θ) = 1.07169 \dots : θ = arctan (1.07169 \dots) + πn

tan (θ) = - 1.65186 \dots : θ = arctan (- 1.65186 \dots) + πn

Combine all the solutions

θ = arctan (- 0.75317 \dots) + πn, θ = arctan (1.07169 \dots) + πn, θ = arctan (- 1.65186 \dots) + πn

Show solutions in decimal form

θ = - 0.64552 \dots + πn, θ = 0.81999 \dots + πn, θ = - 1.02643 \dots + πn

cos (φ) = 3 sin (φ)

5 sin (x) - 3 cos (2 x) + 3 = 0

\frac{tan ^{2} ( x )}{sec ( x ) + 1} = tan (x)

2 cos^{2} (x) + cos (x) = 1, 0 \leq x < 2 π

sin (β) = - 0, 8 (θ \in β v c) s = sec (β) - tan (β)

What is the general solution for tan(2θ)+6tan(θ)+8=0 ?
The general solution for tan(2θ)+6tan(θ)+8=0 is θ=-0.64552…+pin,θ=0.81999…+pin,θ=-1.02643…+pin