Solving for X in a Trig Equation: 5=2sinX + cosX

[21385]

How can you solve for X, an angle, if you get an equation like this?

5=2sinX + cosX

I couldn't think of any trig identities that can solve this, even though this may be extremely easy.

Can someone show me? Thanks

 

[Data]

Well, what you cay say immediately is that the solution is going to be complex. What are the maximum values of sin and cos on the reals?

 

[21385]

srry about that...i just randomly put a few numbers down...

If the equation is like:: 2=3sinX+cosX

How would you solve that?

 

[Eighty]

This one is helpful.
http://en.wikipedia.org/wiki/Trigonometric_identities#Linear_combinations

 

[HallsofIvy]

srry about that...i just randomly put a few numbers down...

If the equation is like:: 2=3sinX+cosX

How would you solve that?

That's a bit complicated but you could do this: Let y= cos(X). Since sin(x)= \sqrt{1- cos^2(x)} we have 2= 3\sqrt{1- y^2}+ y. Rewrite that as 3\sqrt{1- y^2}= 2- y and square both sides: 9(1- y^2)= 4- 4y+ y^2 or 9- 9y^2= 4- 4y+ y^2 so 10y^2- 4y- 5= 0. Solve that using the quadratic formula to find y= cos(X) and take the arcsine to find X.

 

[dimensionless]

Substitute sin(x)= \sqrt{1- cos^2(x)} and solve the quadratic for cos(x). Then take the arcsine of you two solutions. Basically what HallsofIvy said.

 

[daemonakadevil]

or try to substitute :

sin(x) = 2tan(x/2)/1+tan^2(x/2)
and
cos(x)= (1-tan^2(x/2))/1+tan^2(x/2)

it might solve the problem !

 

[drpizza]

Why not get sin on one side and cos on the other side. Square both sides (possibly introducing extraneous roots) and substituting for either sin^2 or cos^2 (using sin^2+cos^2=1). This would result in a quadratic equation. Check the solutions.

 

[danago]

You could write it in the form of 2=Acos(x-\theta)