0
$\begingroup$

Solve the equation

$$3 \sec θ = 4 \cos θ \text{ where ; }0 \leq \theta \leq 90$$

Help find and explain theta.

Thanks.

$\endgroup$
  • 3
    $\begingroup$ Hint: use the fact that $\sec{\theta}=\frac{1}{\cos{\theta}}$. $\endgroup$ – favq Apr 24 '12 at 15:08
  • $\begingroup$ Do you mean to associate this with angles or radians? Angles give $\theta = 30^\circ$ but radians give $\theta = \dfrac{\pi}{6}$ $\endgroup$ – Mr Pie Jan 30 '18 at 10:35
4
$\begingroup$

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).

$\endgroup$
5
$\begingroup$

Hint. $3\sec \theta = 4\cos \theta$ is the same $4 \cos^2\theta - 3 =0$.

$\endgroup$
  • 2
    $\begingroup$ This is true because $\sec{\theta}=\frac{1}{\cos{\theta}}$. Thus, $\frac{3}{\cos{\theta}}=4\cos{\theta}$ $\endgroup$ – favq Apr 24 '12 at 15:13
3
$\begingroup$

$$3 \sec \theta = 4 \cos \theta$$

$$3 \cos \theta \sec \theta = 4 \cos \theta\cos \theta$$

$$3 = 4 \cos ^2 \theta$$

$${3 \over 4} = \cos ^2 \theta$$

$$\pm{\sqrt 3 \over 2} = \cos \theta$$

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 $\pi/6$

$\endgroup$
2
$\begingroup$

$$\frac{3}{\cos \theta}=4 \cos \theta \Rightarrow \frac{4\cos^2 \theta-3}{\cos \theta}=0 \Rightarrow (2\cos \theta-\sqrt 3)(2\cos \theta+\sqrt 3)=0$$

Hence :

$\cos \theta = \frac{-\sqrt 3}{2} ~\text{or}~ \cos \theta =\frac{\sqrt 3}{2}$

Since $\theta$ belongs to the first quadrant we have :

$\theta =\frac{\pi}{6}$

$\endgroup$
2
$\begingroup$

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: $$\cos\theta=\frac{\sqrt{3}}{2}\Leftrightarrow\theta=\frac{\pi}{6}$$

$\endgroup$
  • 2
    $\begingroup$ You could use \cos{blah} and \sin{blah} to get non italic function names $\sin$ and $\cos$, and \left and \right to get delimiters that adapt to their content. compare $$({\sqrt 3 \over 2}-cos\theta)$$ ({\sqrt 3 \over 2}-cos\theta) to $$\left({\sqrt 3 \over 2}-\cos\theta \right)$$ \left({\sqrt 3 \over 2}-\cos\theta \right) $\endgroup$ – Pedro Tamaroff Apr 24 '12 at 15:13
  • $\begingroup$ It is usually a convention to use function names without italics. $\endgroup$ – Pedro Tamaroff Apr 24 '12 at 15:16
  • 2
    $\begingroup$ Ok then, i'll change them!Thanks for the delimiters tip! $\endgroup$ – chemeng Apr 24 '12 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.