Solving the Equation

Question

I need help solving the following equation:

log(base3)(cos2x-tanx+8)=2

 

I'm wondering how to do it, I keep getting stuck after I substitute in for the double angle of cos and am not sure where to go after that.

 

Thank you.

Answer 1

log₃(cos2x - tanx) = 2

 

The solution to a logarithm is the exponent with base number. The base number here is 3

 

 

3² = cos2x - tanx + 8

 

9 = (cos²x - sin²x) - (sinx / cosx) + 8

 

0 = cos²x - sin²x - (sinx / cosx) - 1

 

0 = 1 - sin²x - sin²x - (sinx / cosx) - 1

 

0 = -2sin²x - (sinx / cosx)

0 = (-2sin²x cosx - sinx) / cosx

0 = (sinx (-2sinx cosx - 1)) / cosx

 

Set these equal to zero:

 

sinx = 0                 -2sinx cosx - 1 = 0                 cosx = 0

                             -2sinx cosx = 1                           x = 90

                                sinx cosx = -1/2                       x = 270

x = 0                            x = 60

x = 180                        x = 30

x = 360

 

Our solutions are 0°, 30°, 60°, 180°, and 360°.

Answer 2

Use the log and the equation becomes cos2x + tanx = 0

 

So, cos²x - sin²x + tanx = 0

 

Does this help??  Does using the identities to put this into sin and cos help?

 

Kevin