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

The general solution for (sin(x)-1)(tan(x)+2)=0 is x=-1.10714…+pin

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

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

x = - 1.10714 \dots + πn

Degrees

x = - 63.43494 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Solving each part separately

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

sin (x) - 1 = 0 : x = \frac{π}{2} + 2 πn

sin (x) - 1 = 0

Move

1

to the right side

sin (x) - 1 = 0

Add

1

to both sides

sin (x) - 1 + 1 = 0 + 1

Simplify

sin (x) = 1

sin (x) = 1

General solutions for

sin (x) = 1

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

tan (x) + 2 = 0 : x = arctan (- 2) + πn

tan (x) + 2 = 0

Move

2

to the right side

tan (x) + 2 = 0

Subtract

2

from both sides

tan (x) + 2 - 2 = 0 - 2

Simplify

tan (x) = - 2

tan (x) = - 2

Apply trig inverse properties

tan (x) = - 2

General solutions for

tan (x) = - 2

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

x = arctan (- 2) + πn

x = arctan (- 2) + πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = arctan (- 2) + πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn

x = arctan (- 2) + πn

Show solutions in decimal form

x = - 1.10714 \dots + πn

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

cos^{2} (x) = \frac{( 6 k - 7 )}{7}

(\frac{1 + tanh ( x )}{1 - tanh ( x )})^{2} = 2

cos (2 x) - cos (- x) = 0

cos (3 x) = sin (x)

What is the general solution for (sin(x)-1)(tan(x)+2)=0 ?
The general solution for (sin(x)-1)(tan(x)+2)=0 is x=-1.10714…+pin