If \[ \cos x=\frac45 \] and \(x\) is acute, find \[ \tan2x \]

Solution: Given, \[ \cos x=\frac45 \] Using \[ \sin^2x+\cos^2x=1 \] we get \[ \sin^2x = 1-\left(\frac45\right)^2 \] \[ = 1-\frac{16}{25} \] \[ = \frac9{25} \] \[ \sin x=\pm\frac35 \] Since \(x\) is acute, \[ \sin x>0 \] Therefore, \[ \sin x=\frac35 \] Now, \[ \tan x=\frac{\sin x}{\cos x} \] \[ = \frac{\frac35}{\frac45} \] \[ = \frac34 \] Using the double angle formula: \[ \tan2x=\frac{2\tan x}{1-\tan^2x} \] Substituting the value of \(\tan x\): \[ \tan2x = \frac{2\left(\frac34\right)}{1-\left(\frac34\right)^2} \] \[ = \frac{\frac32}{1-\frac9{16}} \] \[ = \frac{\frac32}{\frac7{16}} \] \[ = \frac32\times\frac{16}{7} \] \[ = \frac{24}{7} \] Hence, \[ \boxed{ \tan2x=\frac{24}{7} } \]