What is the general solution for 5sin(x)tan(x)+4tan(x)=0 ?

The general solution for 5sin(x)tan(x)+4tan(x)=0 is x=pin,x=-0.92729…+2pin,x=pi+0.92729…+2pin

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5sin(x)tan(x)+4tan(x)=0

5 sin (x) tan (x) + 4 tan (x) = 0

x = πn, x = - 0.92729 \dots + 2 πn, x = π + 0.92729 \dots + 2 πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n, x = - 53.13010 \dots^{\circ} + 36 0^{\circ} n, x = 233.13010 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 sin (x) tan (x) + 4 tan (x) = 0

Factor

5 sin (x) tan (x) + 4 tan (x) : tan (x) (5 sin (x) + 4)

5 sin (x) tan (x) + 4 tan (x)

Factor out common term

tan (x)

= tan (x) (5 sin (x) + 4)

tan (x) (5 sin (x) + 4) = 0

Solving each part separately

tan (x) = 0 or 5 sin (x) + 4 = 0

tan (x) = 0 : x = πn

tan (x) = 0

General solutions for

tan (x) = 0

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

5 sin (x) + 4 = 0 : x = arcsin (- \frac{4}{5}) + 2 πn, x = π + arcsin (\frac{4}{5}) + 2 πn

5 sin (x) + 4 = 0

Move

4

to the right side

5 sin (x) + 4 = 0

Subtract

4

from both sides

5 sin (x) + 4 - 4 = 0 - 4

Simplify

5 sin (x) = - 4

5 sin (x) = - 4

Divide both sides by

5

5 sin (x) = - 4

Divide both sides by

5

\frac{5 sin ( x )}{5} = \frac{- 4}{5}

Simplify

sin (x) = - \frac{4}{5}

sin (x) = - \frac{4}{5}

Apply trig inverse properties

sin (x) = - \frac{4}{5}

General solutions for

sin (x) = - \frac{4}{5}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{4}{5}) + 2 πn, x = π + arcsin (\frac{4}{5}) + 2 πn

x = arcsin (- \frac{4}{5}) + 2 πn, x = π + arcsin (\frac{4}{5}) + 2 πn

Combine all the solutions

x = πn, x = arcsin (- \frac{4}{5}) + 2 πn, x = π + arcsin (\frac{4}{5}) + 2 πn

Show solutions in decimal form

x = πn, x = - 0.92729 \dots + 2 πn, x = π + 0.92729 \dots + 2 πn

y, tan (y) = x

2 cos (x) + 5 = 4

cos (x) = - 0.64

y, x = tan (\frac{π}{4} y)

2 sin^{2} (x) - 5 sin (x) + 2 = 0, 0 \leq x \leq 2 π

What is the general solution for 5sin(x)tan(x)+4tan(x)=0 ?
The general solution for 5sin(x)tan(x)+4tan(x)=0 is x=pin,x=-0.92729…+2pin,x=pi+0.92729…+2pin