# Solve that $3 \sec θ = 4 \cos θ$.

Solve the equation

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

Help find and explain theta.

Thanks.

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Hint: use the fact that $\sec{\theta}=\frac{1}{\cos{\theta}}$. –  anonymous Apr 24 '12 at 15:08
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## 5 Answers

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).

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

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

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

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

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

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

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 $\pi/6$

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

Hence :

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

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

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

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

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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) –  Pedro Tamaroff Apr 24 '12 at 15:13
It is usually a convention to use function names without italics. –  Pedro Tamaroff Apr 24 '12 at 15:16
Ok then, i'll change them!Thanks for the delimiters tip! –  chemeng Apr 24 '12 at 15:17
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