simple one i cant figure out! (1 Viewer)

[M-THIS]

hey... it simple... brain needs a kick start!

Q1: COS2X = COSX

Q2: COSX + _/3 SINX = 2 (cos x plus squareroot of 3 sin x)


thanks see if you guyz can help!

 

[Nick]

for the first one, Cos2x equals 2cos squared x

so u make a quadratic, let m equal cosx and solve..

should get zero and 2pi.. i think

 

[wogboy]

1.

cos2x = cosx
2cos^2(x) - 1 = cosx
2cos^2(x) - cosx - 1 = 0
cosx = (1 +- sqrt(9))/4
cosx = 1 or cosx = -1/2
x = 0, x=2*pi, x=2*pi/3, x=4*pi/3 (for 0<=x<=2*pi)

2.

cosx + sqrt(3)*sinx = 2
2(1/2*cosx + (sqrt(3)/2)*sinx) = 2
(1/2)cosx + (sqrt(3)/2)*sinx = 1
cos(pi/3)*cosx + sin(pi/3)*sinx = 1
cos(x - pi/3) = 1
x - pi/3 = 0
x = pi/3 (for 0<=x<=2*pi)

 

[M-THIS]

Thanks WOGBOY your a champ!

so in (2) you can just change the 1/2 and sqrt(3)/2 into their trig. BETWEEN LINE 3 and 4.

and also 5 and 6th lines... you just say that cos = 1 at 0? because (x - pi/3) can't equal 0???

Thanks again!

i got one more if you anyone would like to help!

Q3: sqrt(3) * COSX - SINX = 1 (in general sol.)

this i think is similar to another i can't do : SINX + COSX = 1 again in gen. sol.

Thanks once again...

 

[wogboy]

so in (2) you can just change the 1/2 and sqrt(3)/2 into their trig. BETWEEN LINE 3 and 4.

yep.

and also 5 and 6th lines... you just say that cos = 1 at 0?

yep.

because (x - pi/3) can't equal 0???

it can when x=pi/3

Remember this general method, it's very very important, whenever you see anything in the form:

a*cosx + b*sinx = c, divide both sides of the equation by sqrt(a^2 + b^2). Then try to put it into the form:

cos?cosx + sin?sinx = c/sqrt(a^2 + b^2)

then,

cos(x - ?) = c/sqrt(a^2 + b^2)

If you learn this method, it will never fail you when solving these types of questions

 

[M-THIS]

what you mean this method for Q3... or the previous few?

 

[wogboy]

Use it for *ANY* question in the form:

a*cosx + b*sinx = c (including Q3)

I'll let you do Q3

 

[M-THIS]

so Q3b goes something like this?

COSX + SINX = 1

(COSX + SINX)/sqrt(2) = 1/sqrt(2)

1/sqrt2COSX + 1/sqrt(2)SINX =

TAN30COSX + TAN30SINX = TAN30

than i'm stuck! (i think i did something wrong!)

 

[wogboy]

almost there, but don't use TAN! Use SIN or COS instead. Try again, but this time don't use TAN.

Remember your aim is to get it into the form:

cos?cosx + sin?sinx

not

tan?cosx + tan?sinx

You know this identity? cos(x - y) = cosxcosy + sinxsiny

You need to make use of it. (note that the order of x and y isn't important in cos(x-y), since cos(x-y) = cos(y-x), because cos is an even function).

 

[M-THIS]

thanks man! GOT IT!

owe you one! thanks!

Is this just the way to solve trig eqN?

 

[Lord Ac]

Q3: sqrt(3) * COSX - SINX = 1 (in general sol.)


HOW DO U DO IT!?!?

Ac

 

[Rahul]

auxillary method?
or am i wrong?

 

[Affinity]

sqrt(3) * COSX - SINX = 1

The good method, as mentioned earlier by wogboy:

[sqrt(3) / 2]*cos(x) - [1/2]*sin(x) = 1/2

cos(Pi/6)cos(x) - sin(Pi/6)sin(x) = 1/2

Cos(Pi/6 + x) = 1/2

Pi/6 + x = 2n(Pi) + Pi/3 OR 2n(Pi) - Pi/3

x = 2n(Pi) + Pi/6 OR 2n(Pi) - Pi/2

The Stupid method, if you forgot about the above method:

[1 + sin(x) ] = sqrt(3)*cos(x)
[1 - sin(x) ] = 2 - sqrt(3)*cos(x)

1- [sin(x)]^2 = 2*sqrt(3)*cos(x) - 3[cos(x)]^2

[cos(x)]^2 = 2*sqrt(3)*cos(x) - 3[cos(x)]^2

[cos(x)]^2 -[sqrt(3) / 2]*cos(x) = 0

cos(x)[cos(x) - sqrt(3)/2] = 0

cos(x) = 0 => x = ..., -Pi/2, Pi/2, 3Pi/2, 5Pi/2 ...

cos(x) = sqrt(3)/2 => x=... -Pi/6 , Pi/6, 13Pi/6, 23Pi/6 ...

now testing the solutions in the original equation, eliminating the extra solutions generated by the quadratic:

the solutions are {x| x=-Pi/2 + 2n(Pi) } U {x| x= Pi/6 + 2n(Pi)}

 

[wogboy]

Alternatively, you can use the t-rules if you want to.

sqrt(3)*cosx - sinx = 1

sqrt(3)*(1-t^2)/(1+t^2) - 2t/(1+t^2) = 1
sqrt(3)*(1 - t^2 - 2t) = 1 + t^2

Now solve the quadratic to find t. Also you know that t=tan(x/2), so you can find x.

But of course I prefer the other way I first stated

 

[Affinity]

Thats right!!! the t formula, thanx for reminding me.

 

[Lord Ac]

Thanks, I managed to do it after I slept and could keep my eyes open

Ac