Expert: Paul Klarreich - 2/29/2008

Question
The problem is:  Solve, where 0deg<=x<=360deg.  I need to round the approximate solution(s) to the nearest tenth of a degree.
tan(x)sin(x)-sin(x)=0

Part 2 of my question is i need to solve the equation for solutions in the interval 0<=x<=2pi
cos^2(x)-1=0

If I can get these two examples worked out for me I can complete the rest of what I have to do.  Thank you in advance

Answer
Questioner:   Douglas
Category:  Advanced Math
Private:  No
 
Subject:  Trigonomic equations
Question:  The problem is:  Solve, where 0deg<=x<=360deg.  I need to round the approximate solution(s) to the nearest tenth of a degree.
tan(x)sin(x)-sin(x)=0

Part 2 of my question is i need to solve the equation for solutions in the interval 0<=x<=2pi
cos^2(x)-1=0

If I can get these two examples worked out for me I can complete the rest of what I have to do.  Thank you in advance
..............................
Hi, Doug,

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

This is a 'pseudo-quadratic', and you solve it by factoring:

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

Set sin x = 0,  x = 0,180,[360]

Note: the conditions usually set   0 <= x < 360, not 0 <= x <= 360.

Set tan x - 1 = 0, tan x = 1,  x = 45,225.

............................
 
cos^2(x)-1=0

Note: the conditions usually set   0 <= x < 2pi, not 0 <= x <= 2pi.

Same stuff. But you can also solve:

cos^2(x) = 1
cos x = +- 1.

Set cos x = 1;  x = 0
Set cos x = -1;  x = pi