# Trig: general solution

1. Jun 15, 2009

### DERRAN

1. The problem statement, all variables and given/known data
Given: sec$$\theta$$=$$\sqrt{10}$$ where 0< $$\theta$$ <90

and $$\sqrt{10}$$sin(A-$$\theta$$)=sinA-3cosA

Determine the solution of
6cosA +3 = 2sinA

for A $$\in$$ [-180; 180], rounded off to one decimal digit.

2. Relevant equations

3. The attempt at a solution

3=2sinA - 6cosA

3=2(sinA-3cosA)

$$\frac{3}{2}$$=sinA-3cosA

$$\frac{3}{2}$$=$$\sqrt{10}$$sin(A-$$\theta$$)

Now I can't get rid of the theta

2. Jun 15, 2009

### rock.freak667

if $sec \theta = \sqrt{10}$ that means $cos \theta = ?$

3. Jun 15, 2009

### DERRAN

cos$$\theta$$=1/$$\sqrt{10}$$
but that still doen't help with getting rid of the theta over here.

3/2=$$\sqrt{10}$$sin(A-$$\theta$$)

4. Jun 15, 2009

### rock.freak667

erm...

$$cos \theta = \frac{1}{\sqrt{10}}$$

$\theta$ is what then in the range $0< \theta < 90$ ?

5. Jun 15, 2009

### DERRAN

got it Thank you.