What is the general solution for (1+tan(x))(5sin(x)-2)=0 ?

The general solution for (1+tan(x))(5sin(x)-2)=0 is x=(3pi)/4+pin,x=0.41151…+2pin,x=pi-0.41151…+2pin

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(1+tan(x))(5sin(x)-2)=0

(1 + tan (x)) (5 sin (x) - 2) = 0

x = \frac{3 π}{4} + πn, x = 0.41151 \dots + 2 πn, x = π - 0.41151 \dots + 2 πn

Solution steps

(1 + tan (x)) (5 sin (x) - 2) = 0

Solving each part separately

1 + tan (x) = 0 or 5 sin (x) - 2 = 0

1 + tan (x) = 0 : x = \frac{3 π}{4} + πn

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

Combine all the solutions

x = \frac{3 π}{4} + πn, x = arcsin (\frac{2}{5}) + 2 πn, x = π - arcsin (\frac{2}{5}) + 2 πn

Show solutions in decimal form

x = \frac{3 π}{4} + πn, x = 0.41151 \dots + 2 πn, x = π - 0.41151 \dots + 2 πn

arccos (e - x) = π^{3}

sin (a) = \frac{20}{25}

1 + cos (x) = 3 - sin (x)

3 = 7 cos (\frac{π}{3} t)

cos^{2} (2 x) = cos (4 x)

What is the general solution for (1+tan(x))(5sin(x)-2)=0 ?
The general solution for (1+tan(x))(5sin(x)-2)=0 is x=(3pi)/4+pin,x=0.41151…+2pin,x=pi-0.41151…+2pin