What is the general solution for tan(3φ-4)=-1/2 ,0<= φ<= pi/2 ?

The general solution for tan(3φ-4)=-1/2 ,0<= φ<= pi/2 is φ=(-0.46364…+4)/3 ,φ=(-0.46364…-pi+4)/3

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tan(3φ-4)=-1/2 ,0<= φ<= pi/2

tan (3 φ - 4) = - \frac{1}{2}, 0 \leq φ \leq \frac{π}{2}

φ = \frac{- 0.46364 \dots + 4}{3}, φ = \frac{- 0.46364 \dots - π + 4}{3}

Solution steps

tan (3 φ - 4) = - \frac{1}{2}, 0 \leq φ \leq \frac{π}{2}

Apply trig inverse properties

3 φ - 4 = arctan (- \frac{1}{2}) + πn

Solve

3 φ - 4 = arctan (- \frac{1}{2}) + πn : φ = - \frac{arctan ( \frac{1}{2} )}{3} + \frac{πn}{3} + \frac{4}{3}

φ = - \frac{arctan ( \frac{1}{2} )}{3} + \frac{πn}{3} + \frac{4}{3}

Solutions for the range

0 \leq φ \leq \frac{π}{2}

φ = \frac{- arctan ( \frac{1}{2} ) + 4}{3}, φ = \frac{- arctan ( \frac{1}{2} ) - π + 4}{3}

Show solutions in decimal form

φ = \frac{- 0.46364 \dots + 4}{3}, φ = \frac{- 0.46364 \dots - π + 4}{3}

g, θ = arctan (\frac{5.54}{10 g} + 0.2)

tan (t) = \frac{3}{4}

2 cos (θ) - 2 sin (2 θ) = 0

tan (x) = \frac{5}{3} sin (x) [0.2 π]

4 sin^{2} (x) - cos^{2} (x) = 0

What is the general solution for tan(3φ-4)=-1/2 ,0<= φ<= pi/2 ?
The general solution for tan(3φ-4)=-1/2 ,0<= φ<= pi/2 is φ=(-0.46364…+4)/3 ,φ=(-0.46364…-pi+4)/3