Trig - Addition Formula Question

[studentxlol]

Homework Statement

Solve the equation cos(x-60)=sinx

Homework Equations

cosAcosB+sinAsinB

The Attempt at a Solution

cos(x-60)=sinx
cosxcos60+sinxsin60=sinx
1/2cosx+(√3)/2sinx=sinx

How do I then solve to find x for0<x<360

 

[SammyS]

Homework Statement

Solve the equation cos(x-60)=sinx

Homework Equations

cosAcosB+sinAsinB

The Attempt at a Solution

cos(x-60)=sinx
cosxcos60+sinxsin60=sinx
1/2cosx+(√3)/2sinx=sinx

How do I then solve to find x for0<x<360

Subtract sin(x) from both sides.

Divide by cos(x).

 

[Curious3141]

Homework Statement

Solve the equation cos(x-60)=sinx

Homework Equations

cosAcosB+sinAsinB

The Attempt at a Solution

cos(x-60)=sinx
cosxcos60+sinxsin60=sinx
1/2cosx+(√3)/2sinx=sinx

How do I then solve to find x for0<x<360

Suggestion: don't even do it this way.

You know that sin x = cos (90-x)

So re-express RHS like that:

cos (x-60) = cos (90-x)

x-60 = 90-x + 360n (where n is an integer)

2x = 150 + 360n

x = 75 + 180n

So x = 75 or 255 for n = 0 and 1 respectively. Those are the only two solutions in the required range.

 

[SammyS]

Homework Statement

Solve the equation cos(x-60)=sinx

Homework Equations

cosAcosB+sinAsinB

The Attempt at a Solution

cos(x-60)=sinx
cosxcos60+sinxsin60=sinx
1/2cosx+(√3)/2sinx=sinx

How do I then solve to find x for0<x<360

Subtract sin(x) from both sides.

Divide by cos(x).

Did you try what I suggested ?

What did you get ?

(1/2)cos(x)+((√3)/2-1)sin(x)=0

Now, divide by cos(x) .

\displaystyle \frac{\sin(x)}{\cos(x)}=\tan(x) ---- Right?