## sabika13 Group Title solve for x in the given interval 2tanx=secx (-2pi,0) one year ago one year ago

$$\large 2tanx=secx$$ $$\large 2\frac{sinx}{cosx}=\frac{1}{cosx}$$ $$\large \frac{2sinx}{cosx}=\frac{1}{cosx}$$ $$\large 2sinx=1$$ numerators must be equal $$\large sinx=\frac{1}{2}$$ Can you do the rest from here?