# Solve another trig eqn with square roots

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 dont know what to do after how do I get everything on the right hand side and simplify it?

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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?

I need to solve it yes u are right i cant use the trig identity 1+sin^2 theta to replace cos... now im really lost lol what do I do? This is a review question Hurkyl
Staff Emeritus
Gold Member
If you want to have a cos2 &theta;, then simply manipulate the equation until you get a cos2 &theta;.

I dont want to get cos^2 theta i just thought that was the way to solve this but I really have no idea Help plz

Hurkyl
Staff Emeritus
Gold Member
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
aisha said:
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 dont 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:

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

aisha said:
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.

Ouabache
Homework Helper
xanthym said:
√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:
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, :surprised
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
Homework Helper
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.

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