1. ## Trig solution

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

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

Determine the solution of
6cosA +3 = 2sinA

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

2. Relevant equations

3. The attempt at a solution

3=2sinA - 6cosA

3=2(sinA-3cosA)

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

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

Now I can't get rid of the theta

2. ## [MATH]\theta[/MATH]

Good evening

If $\displaystyle sec\theta=\sqrt{10}$, then

$\displaystyle \sqrt{10}sin(A-\theta)=\frac{sin(A-\theta)}{cos\theta}$
$\displaystyle \frac{sinAcos \theta-sin\theta cosA}{cos\theta}$
$\displaystyle sinA-tan\theta cosA=sinA-3cosA$

Hence $\displaystyle tan\theta=3$
$\displaystyle \theta=tan^{-1}3$

Convert to:$\displaystyle \frac{3}{2\sqrt{10}}=sin(A-(tan^{-1}3))$