Question

Solve the equation 2cosA = 3tanA

Answer 1

$\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta) \\\\\\ tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\\\\ -----------------------------\\\\ 2cos(A)=3tan(A)\implies 2cos(A)=3\cfrac{sin(A)}{cos(A)} \\\\\\ 2cos^2(A)=3sin(A)\implies 2[1-sin^2(A)]=3sin(A) \\\\\\ 2-2sin^2(A)=3sin(A)\implies 2sin^2(A)+3sin(A)-2$

$\bf \\\\\\ 0=[2sin(A)-1][sin(A)+2]\implies \begin{cases} 0=2sin(A)-1\\ 1=2sin(A)\\ \frac{1}{2}=sin(A)\\\\ sin^{-1}\left( \frac{1}{2} \right)=\measuredangle A\\\\ \frac{\pi }{6},\frac{5\pi }{6}\\ ----------\\ 0=sin(A)+2\\ -2=sin(A) \end{cases}$

now, as far as the second case....well, sine of anything is within the range of -1 or 1, so -1 < sin(A) < 1

now, we have -2 = sin(A), which simply is out of range for a valid sine, so there's no angle with such sine

so, only the first case are the valid angles for A