Trig: Solving 6cosA+3=2sinA with Given Conditions

[DERRAN]

Homework Statement

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.

Homework Equations

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

 

[rock.freak667]

Homework Statement

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.

Homework Equations

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

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

 

[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)

 

[rock.freak667]

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)

erm...


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

\theta is what then in the range 0&lt; \theta &lt; 90 ?

 

[DERRAN]

erm...


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

\theta is what then in the range 0&lt; \theta &lt; 90 ?

got it Thank you.