Solve another trig eqn with square roots

[aisha]

solve $$\sqrt2 \sin\theta= \sqrt3-\cos\theta$$ algebraically for the domain 0<theta<2pi

I know that the cos can be changed into 1-sin^2 theta but I don't know what to do after how do I get everything on the right hand side and simplify it?

 

[ToxicBug]

That cos can't be changed into what you say since its not cos². Anyway, what exactly are you supposed to do with that equation?

 

[aisha]

I need to solve it yes u are right i can't use the trig identity 1+sin^2 theta to replace cos... now I am really lost lol what do I do? This is a review question

 

[Hurkyl]

If you want to have a cos² &theta;, then simply manipulate the equation until you get a cos² &theta;.

 

[aisha]

I don't want to get cos^2 theta i just thought that was the way to solve this but I really have no idea Help please

 

[Hurkyl]

The way I prefer to solve this type of problem is to rearrange it to look like:

A sin x + B cos x = C

So that I may find a y such that

A = sin y
B = cos y

which allows me to use the addition of angles formula for sine.

(Note that you need to have A^2 + B^2 = 1 to be able to do this... so you may need to multiply through by a constant so that this will be true)

 

[xanthym]

solve $$\sqrt2 \sin\theta= \sqrt3-\cos\theta$$ algebraically for the domain 0<theta<2pi

I know that the cos can be changed into 1-sin^2 theta but I don't know what to do after how do I get everything on the right hand side and simplify it?

√2*sin(θ) = √3 - cos(θ)
⇒ √2*{1 - cos^2(θ)}^(1/2) = √3 - cos(θ) (<--- sin^2() + cos^2() = 1)
⇒ 2*{1 - cos^2(θ)} = 3 - 2*√3*cos(θ) + cos^2(θ) (<--- Squaring Both Sides)
⇒ 3*cos^2(θ) - 2*√3*cos(θ) + 1 = 0
⇒ {cos(θ) - 1/√3}^2 = 0
⇒ θ = arccos(1/√3)
⇒ Possible solutions: θ = {(0.9553 rad)=(54.736 deg)} or {(5.328 rad)=(305.264 deg)}
Checking original equation, only first possibility is solution:
θ = {(0.9553 rad)=(54.736 deg)}


~~

 

[aisha]

the solutions are 55 degrees and 305 degrees I found the last post to be a little confusing

 

[The Bob]

the solutions are 55 degrees and 305 degrees I found the last post to be a little confusing

If you do understand the last post then all you need to do is take the 55° and then minus it from 360° and then you will end up with both answers.

The Bob (2004 ©)

 

[Ouabache]

√2*sin(θ) = √3 - cos(θ)
⇒ √2*{1 - cos^2(θ)}^(1/2) = √3 - cos(θ) (<--- sin^2() + cos^2() = 1)
⇒ 2*{1 - cos^2(θ)} = 3 - 2*√3*cos(θ) + cos^2(θ) (<--- Squaring Both Sides)
⇒ 3*cos^2(θ) - 2*√3*cos(θ) + 1 = 0
⇒ {cos(θ) - 1/√3}^2 = 0
⇒ θ = arccos(1/√3)
⇒ Possible solutions: θ = {(0.9553 rad)=(54.736 deg)} or {(5.328 rad)=(305.264 deg)}
Checking original equation, only first possibility is solution:
θ = {(0.9553 rad)=(54.736 deg)}

Where did you become confused? Did you understand the substitution xanthym made for $$\sin\theta$$ ?

$$\sin\theta = [1 - cos^2\theta]^\frac{1}{2}$$

With that substitution, the equation became in terms of $$\cos\theta$$.
After squaring both sides and rearranging terms, xanthym rewrote the equation so it looks like:
$$ax^2+bx+c = 0$$ ... $$3cos^2\theta - 2\sqrt{3}\cos\theta + 1 = 0$$

Can you factor that equation directly?
If that looks too tricky,
do you know another formula you can use to find roots of an equation in that form?

Do you understand taking the arccos (inverse cosine) of a value to find the angle?

(hint: to find both angles, remember there will be more than one quadrant between 0 and 2pi, where cosθ has the same sign).

 

[dextercioby]

xanthym got his one right.That 305° is not a viable solution,bacause it does not satisfy the initial equation.The sine of it is negative,while the the sqrt{3}-\cosine of it is positive...

Daniel.

 

[aisha]

Thanks it took me a little time to absorb but i do understand

 

[dextercioby]

38hrs...That's reasonable...You know what they say:"better late than never".Of course,it does not apply to dating women...

Daniel.