Trig question

1. Apr 8, 2007

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

2. Apr 8, 2007

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?

3. Apr 9, 2007

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?

4. Apr 9, 2007

Eighty

5. Apr 10, 2007

HallsofIvy

Staff Emeritus
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.

6. Apr 10, 2007

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.

7. Apr 10, 2007

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 !

8. Apr 10, 2007

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.

9. Apr 10, 2007

danago

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