-----------> general solutions???....trig.....arrgh (1 Viewer)

[lillaila]

Can n e one work this out, plz?
1. sinx + cosx = 1
2. (sq.root 3)cosx - sinx = 1

The answers are:
1. x = 2npi or 2npi + (pi on 2)
* i worked it out with "sin"....but the back of the book used "cos", hence 2npi ( ??? )
I got (for "sin") x = npi or npi + [(-1)^n](pi on 2)
is this right?

2. x = 2npi + (pi on 6) or 2npi - (pi on 2)
* shouldn't "cos" have + OR - ? i don't understand y they left out one or the other.

Thnku!

 

[NSBHSchoolie]

1. sinx+cosx=1
(squaring both sides)
sin^2x+2sinxcosx+cos^2x=1
(sin^2x+cos^2x=1)
2sinxcosx=0
sin2x=0 (lets say the domain is 0<x<2pi)
sinx=0 for 0<2x<4pi
2x=0, pi, 2pi, 3pi, 4pi.
x=0, pi/2, pi, 3pi/2, 2pi, for 0<x<2pi

however pi and 3pi/2 are not solutions, as they are negative.

x=0, pi/2, 2pi

hope thats right...anyone else?

That second one's hard.

 

[McLake]

NSBHSchoolie's response looks right for 1

----------------------------------------------------

2.
sqrt(3)*cosx - sinx = 1

()^2 both sides

3cos^2x - 2sqrt(3)*sinxcosx + sin^2x = 1

now cos^2x + sin^2x = 1

2cos^2x - 2sqrt(3)*sinxcosx = 0

/2

cos^2x - sqrt(3)*sinxcosx = 0

now heres the tricky part:

/cos^2x (cos^2x != 0)

1 - sqrt(3)*sinx/cosx = 0
1 = sqrt(3)*tanx
tanx = 1/sqrt(3)
x= 2npi +/- (pi/6)

But now we have to check that cos^2x != 0

If cos^2x = 0
Then cosx = 0
Then sinx = 1
Then x = pi/2

so x can also be 2npi +/- pi/2. Um, that's not quite your answer though ...

 

[OLDMAN]

Sorry kids, can't just sit here and watch.

There are two recommended approach to problems of type,

acos@+bsin@=c

Approach 1: "t" method
Approach 2: angle sum method, sin(alpha+theta)=C

Both approaches are well documented in any 3-unit book.

 

[lillaila]

Thanku pple ...