Solve the equation
$3 \sec θ = 4 \cos θ \text{ where ; }0 \leq \theta \leq 90$
Help find and explain theta.
Thanks.
Solve the equation
$3 \sec θ = 4 \cos θ \text{ where ; }0 \leq \theta \leq 90$
Help find and explain theta.
Thanks.
$\frac{3}{\cos \theta}=4 \cos \theta \Rightarrow \frac{4\cos^2 \theta-3}{\cos \theta}=0 \Rightarrow (2\cos \theta-\sqrt 3)(2\cos \theta+\sqrt 3)=0$
Hence :
$\cos \theta = \frac{-\sqrt 3}{2} ~\text{or}~ \cos \theta =\frac{\sqrt 3}{2}$
Since $\theta$ belongs to the first quadrant we have :
$\theta =\frac{\pi}{6}$
Hint. $3\sec \theta = 4\cos \theta$ is the same $4 \cos^2\theta - 3 =0$.
You start with the equation $\tag{1} 3\sec\theta =4\cos\theta. $ Since $\sec\theta={1\over \cos\theta}$, equation $(1)$ is equivalent to the equation $ \tag{2} {3\over\cos\theta}=4\cos\theta. $ By "equivalent", I mean that $\theta$ is a solution of equation $(1)$ if and only if it is a solution of equation $(2)$. It seems that multiplying both sides of $(2)$ by $\cos\theta$ would simplify matters. This would give $\tag{3} 3=4\cos^2\theta $ But wait, is equation $(3)$ equivalent to equation $(2)$? We need to be concerned with what happens if $\cos\theta=0$. Here the left hand side of equation $(2)$ is not defined when $\cos\theta=0$. On the other hand $\cos\theta=0$ does not lead to a solution of equation $(3)$. So, indeed, equations $(2)$ and $(3)$ are equivalent: multiplication by $\cos\theta$ is valid when it is not zero, and when $\cos\theta=0$, neither $(2)$ nor $(3)$ have a solution.
Ok then, we need to solve $(3)$. Dividing both sides by $4$ gives $\tag{4} \cos^2\theta={3\over4}. $ Taking the square roots of both sides gives $ |\cos\theta|=\sqrt3/2. $ Since we know $0^\circ\le\theta\le90^\circ$, we know that $|\cos\theta|=\cos\theta$ (the cosine function is nonnegative in the first quadrant), so we have to find the value(s) of $\theta$ in $[0^\circ,90^\circ]$ for which $ \cos\theta =\sqrt3/2. $
You should recognize that there is only one solution here, namely $\theta=30^\circ$ (you're dealing with one of the "special angles", here).
$3 \sec \theta = 4 \cos \theta$
$3 \cos \theta \sec \theta = 4 \cos \theta\cos \theta$
$3 = 4 \cos ^2 \theta$
${3 \over 4} = \cos ^2 \theta$
$\pm{\sqrt 3 \over 2} = \cos \theta$
The solutions are all angles related to $\pi/6$ and $11\pi/6$ for the positive solution, and to $5\pi/6$ and $7\pi/6$ for the negative solution, by multiples of $2\pi$.
In your case you only want $\pi/6$
You got: $\sec\theta=\frac{1}{\cos\theta}$ therefore your equation becomes: $\frac{3}{4}=\cos^2\theta\Leftrightarrow\frac{3}{4}-\cos^2\theta=0\Leftrightarrow\left(\frac{\sqrt{3}}{2}-\cos\theta\right)\left(\frac{\sqrt{3}}{2}+\cos\theta\right) =0$Since $0 \leq \theta \leq 90$, $\cos\theta$ is positive, hence: $\cos\theta=\frac{\sqrt{3}}{2}\Leftrightarrow\theta=\frac{\pi}{6}$