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

The general solution for 2sin(x)tan(x)-tan(x)-2sin(x)+1=0 is x= pi/6+2pin,x=(5pi)/6+2pin,x= pi/4+pin

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Popular >

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

Solution

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

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{π}{4} + πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Factor

2 sin (x) tan (x) - tan (x) - 2 sin (x) + 1 : (2 sin (x) - 1) (tan (x) - 1)

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

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

Factor out

- 1

from

- 2 sin (x) + 1 : - (2 sin (x) - 1)

- 2 sin (x) + 1

Factor out common term

- 1

= - (2 sin (x) - 1)

Factor out

tan (x)

from

2 sin (x) tan (x) - tan (x) : tan (x) (2 sin (x) - 1)

2 sin (x) tan (x) - tan (x)

Factor out common term

tan (x)

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

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

Factor out common term

2 sin (x) - 1

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

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

Solving each part separately

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

2 sin (x) - 1 = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

2 sin (x) - 1 = 0

Move

1

to the right side

2 sin (x) - 1 = 0

Add

1

to both sides

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

Simplify

2 sin (x) = 1

2 sin (x) = 1

Divide both sides by

2

2 sin (x) = 1

Divide both sides by

2

\frac{2 sin ( x )}{2} = \frac{1}{2}

Simplify

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

General solutions for

sin (x) = \frac{1}{2}

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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

tan (x) - 1 = 0

Move

1

to the right side

tan (x) - 1 = 0

Add

1

to both sides

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

Simplify

tan (x) = 1

tan (x) = 1

General solutions for

tan (x) = 1

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 = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{π}{4} + πn

Popular Examples

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sin (2 x) = \frac{( 6 m - 5 )}{5}

4cos^2(x)+6=7

4 cos^{2} (x) + 6 = 7

sec(θ)-1=(sqrt(2)-1)tan(θ)

sec (θ) - 1 = (2 - 1) tan (θ)

cos(a)=-4/5

cos (a) = - \frac{4}{5}

2tan(θ)-sec(θ)=0

2 tan (θ) - sec (θ) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(x)tan(x)-tan(x)-2sin(x)+1=0 ?
The general solution for 2sin(x)tan(x)-tan(x)-2sin(x)+1=0 is x= pi/6+2pin,x=(5pi)/6+2pin,x= pi/4+pin