Welcome :: Homework Help and Answers :: Mathskey.com

Recent Visits

  
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

12,761 questions

17,393 answers

1,428 comments

33,429 users

Solve the equation for solutions over the interval [0, 360)?

0 votes
hese just totally lose me :(

I'll use U for theta.

2 tan^2 U cos U - tan^2 U =0

thank you!!
asked Apr 17, 2013 in TRIGONOMETRY by homeworkhelp Mentor

1 Answer

0 votes

2tan2U cosU - tan2U = 0 [0 , 360)

Take out common term tan2U

tan2U(2cosU - 1) = 0

Tan2U = 0 and 2cosU - 1 = 0

if tan2U = 0 then tanU = 0 or U = 0

2cosU - 1 = 0

2cosU = 1

cosU = 1 / 2

Therefore U = π/ 3 , 5π / 3(60 , 300)

The equation for solutions over the interval [0 , 360) :U = 0 , 60, 300 or 0 , π / 3 , 5π / 3.

 

answered Apr 18, 2013 by diane Scholar

 

  • tan U = 0.

tan U = tan 0.

The genaral solution of tan(θ) = tan(α) is θ = nπ + α, where n is an integer.

⇒ U = nπ + 0

U = .

If n = 0, U = (0)π = 0,

If n = 1, U = (1)π = π,

If n = 2, U = (2)π= .

U = 0 and U = π in the interval [0, 2π).

  • cos U = 1/2

cos (U) = cos(π/3)

The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

⇒U = 2nπ ± (π/3)

If n = 0, U = 2(0)π + (π/3) and U = 2(0)π - (π/3) = π/3 and - π/3,

If n = 1, U = 2(1)π + (π/3) and U = 2(1)π - (π/3) = 2π + π/3 and 2π - π/3 = 7π/3 and 5π/3.

U = π/3 and U = 5π/3 in the interval [0, 2π).

Therefore, the solutions of the given equation are U = 0, U = π/3, U = π and U = 5π/3 in the interval [0, 2π).

 

Related questions

...