Solve for x between 0 , and 2pi the next equation?

[luigihs]

Solve for x between 0 , and 2pi the next equation?

cos x + square root(3) sinx = square root(2)

sin(x+ (pi/6)) = 1/sqrt2 <-- but then I don't know what to do this what I think should do, but I am not sure how to solve x help me thanks!

 

[jedishrfu]

You need to present your homework using the homework template where you state the problem, show some relevant formula that you might use to solve it and show an attempt.

I don't see how you got the sin(x+pi/6)=1/sqr(2) result from your original equation.

 

[Mark44]

Solve for x between 0 , and 2pi the next equation?

cos x + square root(3) sinx = square root(2)

sin(x+ (pi/6)) = 1/sqrt2 <-- but then I don't know what to do this what I think should do, but I am not sure how to solve x help me thanks!

So sin(x + π/6) = √2/2

Can you solve sin(u) = √2/2?

You've done all the hard work. The rest is easy.

 

[luigihs]

You need to present your homework using the homework template where you state the problem, show some relevant formula that you might use to solve it and show an attempt.

I don't see how you got the sin(x+pi/6)=1/sqr(2) result from your original equation.

Im really confused I don't even know any relevant formula I just tried to move the x

 

[luigihs]

So sin(x + π/6) = √2/2

Can you solve sin(u) = √2/2?

You've done all the hard work. The rest is easy.

is this right? because somebody said early before that I was wrong... sin (u) = 0.707 ?

 

[Mark44]

You need to present your homework using the homework template where you state the problem, show some relevant formula that you might use to solve it and show an attempt.

Good advice. luigihs, the template is there for a reason - don't just blow it away...

I don't see how you got the sin(x+pi/6)=1/sqr(2) result from your original equation.

He divided both sides of the equation by 2 to get (1/2)sin(x) + (√3/2)cos(x) = √2/2, and then replaced 1/2 by cos(π/6) and √3/2 by sin(π/6). It would have been helpful if he had shown his steps.

 

[luigihs]

Sorry for what is the answer 0.707 ? I don't know how to do the operation

 

[jedishrfu]

Good advice. luigihs, the template is there for a reason - don't just blow it away...

He divided both sides of the equation by 2 to get (1/2)sin(x) + (√3/2)cos(x) = √2/2, and then replaced 1/2 by cos(π/6) and √3/2 by sin(π/6). It would have been helpful if he had shown his steps.

Thanks I didn't see that. I kept thinking to square both sides but then I was still left with a puzzle. Its been a long time since I played with trig problems.

 

[SammyS]

@ luigihs,

If \displaystyle \ \sin(\theta)=\frac{\sqrt{2}}{2}\,,\ then what is θ ?