What is the general solution for (tan(θ)-1)(sec(θ)-1)=0,0<= θ<= 2pi ?

Q: What is the general solution for (tan(θ)-1)(sec(θ)-1)=0,0<= θ<= 2pi ?

The general solution for (tan(θ)-1)(sec(θ)-1)=0,0<= θ<= 2pi is θ= pi/4 ,θ=(5pi)/4 ,θ=0,θ=2pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(tan(θ)-1)(sec(θ)-1)=0,0<= θ<= 2pi

Solution

(tan (θ) - 1) (sec (θ) - 1) = 0, 0 \leq θ \leq 2 π

Solution

θ = \frac{π}{4}, θ = \frac{5 π}{4}, θ = 0, θ = 2 π

Degrees

θ = 4 5^{\circ}, θ = 22 5^{\circ}, θ = 0^{\circ}, θ = 36 0^{\circ}

Solution steps

(tan (θ) - 1) (sec (θ) - 1) = 0, 0 \leq θ \leq 2 π

Solving each part separately

tan (θ) - 1 = 0 or sec (θ) - 1 = 0

tan (θ) - 1 = 0, 0 \leq θ \leq 2 π : θ = \frac{π}{4}, θ = \frac{5 π}{4}

tan (θ) - 1 = 0, 0 \leq θ \leq 2 π

Move

1

to the right side

tan (θ) - 1 = 0

Add

1

to both sides

tan (θ) - 1 + 1 = 0 + 1

Simplify

tan (θ) = 1

tan (θ) = 1

General solutions for

tan (θ) = 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}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{4}, θ = \frac{5 π}{4}

sec (θ) - 1 = 0, 0 \leq θ \leq 2 π : θ = 0, θ = 2 π

sec (θ) - 1 = 0, 0 \leq θ \leq 2 π

Move

1

to the right side

sec (θ) - 1 = 0

Add

1

to both sides

sec (θ) - 1 + 1 = 0 + 1

Simplify

sec (θ) = 1

sec (θ) = 1

General solutions for

sec (θ) = 1

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = 0, θ = 2 π

Combine all the solutions

θ = \frac{π}{4}, θ = \frac{5 π}{4}, θ = 0, θ = 2 π

Graph

Popular Examples

cos(θ)=(4.4)/(11)

cos (θ) = \frac{4.4}{11}

solvefor x,cos^3(x)-cos(x)=0 solve for

x, cos^{3} (x) - cos (x) = 0

-6sin^2(θ)-3=sin(θ)-5

- 6 sin^{2} (θ) - 3 = sin (θ) - 5

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

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

2sqrt(3)sin(2x)-3=0

23 sin (2 x) - 3 = 0

Frequently Asked Questions (FAQ)

What is the general solution for (tan(θ)-1)(sec(θ)-1)=0,0<= θ<= 2pi ?
The general solution for (tan(θ)-1)(sec(θ)-1)=0,0<= θ<= 2pi is θ= pi/4 ,θ=(5pi)/4 ,θ=0,θ=2pi