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

The general solution for tan(α)=-4/3 , pi/2 <α<pi,cos(β)= 1/2 ,0<β< pi/2 is α=-0.92729…+pi

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tan(α)=-4/3 , pi/2 <α<pi,cos(β)= 1/2 ,0<β< pi/2

tan (α) = - \frac{4}{3}, \frac{π}{2} < α < π, cos (β) = \frac{1}{2}, 0 < β < \frac{π}{2}

α = - 0.92729 \dots + π

Degrees

α = 126.86989 \dots^{\circ}

Solution steps

tan (α) = - \frac{4}{3}, \frac{π}{2} < α < π, cos (β) = \frac{1}{2}, 0 < β < \frac{π}{2}

Apply trig inverse properties

tan (α) = - \frac{4}{3}

General solutions for

tan (α) = - \frac{4}{3}

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

α = arctan (- \frac{4}{3}) + πn

α = arctan (- \frac{4}{3}) + πn

Solutions for the range

\frac{π}{2} < α < π

α = - arctan (\frac{4}{3}) + π

Show solutions in decimal form

α = - 0.92729 \dots + π

- 3 cot (\frac{3 x}{2}) csc (\frac{3 x}{2}) = 0

sec (θ) = \frac{17}{8}

3 sin^{2} (x) = 1

3 sin (θ) + 5 = - 2 sin (θ)

2 - 4 sin (θ) = 0

What is the general solution for tan(α)=-4/3 , pi/2 <α<pi,cos(β)= 1/2 ,0<β< pi/2 ?
The general solution for tan(α)=-4/3 , pi/2 <α<pi,cos(β)= 1/2 ,0<β< pi/2 is α=-0.92729…+pi