Find the Variables

Problem

[[cos(3*x),sin(3*x)],[−sin(3*x),cos(3*x)]]*[[b],[d]]=[[0],[tan(3*x)]]

Solution

1. Perform matrix multiplication on the left side to create a system of linear equations.

b*cos(3*x)+d(sin(3*x))=0

−b*sin(3*x)+d(cos(3*x))=tan(3*x)

2. Solve for b in the first equation by isolating the term.

b*cos(3*x)=−d(sin(3*x))

b=−d(sin(3*x)/cos(3*x))

b=−d(tan(3*x))

3. Substitute the expression for b into the second equation.

−(−d(tan(3*x)))*sin(3*x)+d(cos(3*x))=tan(3*x)

d(tan(3*x))*sin(3*x)+d(cos(3*x))=tan(3*x)

4. Factor out d and simplify the trigonometric expression using the identity tan(3*x)=sin(3*x)/cos(3*x)

d*(sin(3*x)/cos(3*x)*sin(3*x)+cos(3*x))=tan(3*x)

d((sin2(3*x)+cos2(3*x))/cos(3*x))=tan(3*x)

5. Apply the Pythagorean identity sin2(3*x)+cos2(3*x)=1 to solve for d

d(1/cos(3*x))=tan(3*x)

d=tan(3*x)*cos(3*x)

d=sin(3*x)/cos(3*x)*cos(3*x)

d=sin(3*x)

6. Substitute d back into the equation for b

b=−(sin(3*x))*tan(3*x)

b=−sin(3*x)sin(3*x)/cos(3*x)

b=−sin2(3*x)/cos(3*x)

Final Answer

b=−sin2(3*x)/cos(3*x),d=sin(3*x)

Want more problems? Check here!