# Given that 7tan a + 2 cos a= 5sec a,derive a quadratic equation for sin a. Hence, find all values of a in the interval [0,pi] which satisfy the given equation?

Aug 29, 2015

$a = \frac{\pi}{6} \text{ }$ or $\text{ } a = {30}^{\circ}$

#### Explanation:

You will use three trigonometric identities to solve this equation

$\tan x = \sin \frac{x}{\cos} x \text{ }$, $\text{ } \sec x = \frac{1}{\cos} x$

and

${\cos}^{2} x + {\sin}^{2} x = 1$

Start by replacing $\tan \left(a\right)$ and $\sec \left(a\right)$ into the original equation

$7 \cdot \sin \frac{a}{\cos} \left(a\right) + 2 \cos \left(a\right) = 5 \cdot \frac{1}{\cos} \left(a\right)$

Right from the start, any value of $a$ that would make $\cos \left(a\right) = 0$ will be excluded from the solutions set.

For the interval $\left[0 , \pi\right]$, $\cos \left(a\right) = 0$ for $a = \frac{\pi}{2}$.

So, multiply $2 \cos \left(a\right)$ by $1 = \cos \frac{a}{\cos} \left(a\right)$ to get rid of the denominators

$7 \cdot \sin \frac{a}{\cos} \left(a\right) + 2 {\cos}^{2} \frac{a}{\cos} \left(a\right) = 5 \cdot \frac{1}{\cos} \left(a\right)$

This is equivalent to

$7 \cdot \sin \left(a\right) + 2 {\cos}^{2} \left(a\right) - 5 = 0$

Use ${\cos}^{2} \left(a\right) = 1 - {\sin}^{2} \left(a\right)$ to get

$7 \cdot \sin \left(a\right) + 2 \left[1 - {\sin}^{2} \left(a\right)\right] - 5 = 0$

The quadratic form of this equaation will thus be

$- 2 {\sin}^{2} \left(a\right) + 7 \sin \left(a\right) - 3 = 0$

Use the quadratic formula to get the two roots of the quadratic

${\sin}_{1 , 2} \left(a\right) = \frac{- 7 \pm \sqrt{{7}^{2} - 4 \cdot \left(- 2\right) \cdot \left(- 3\right)}}{2 \cdot \left(- 2\right)}$

${\sin}_{1 , 2} \left(a\right) = \frac{- 7 \pm \sqrt{25}}{\left(- 4\right)} = \frac{- 7 \pm 5}{\left(- 4\right)}$

The two roos will thus be

$\textcolor{red}{\cancel{\textcolor{b l a c k}{{\sin}_{1} \left(a\right) = \frac{- 7 - 5}{\left(- 4\right)} = 3}}} \text{ }$ and $\text{ "sin_2(a)= (-7 + 5)/((-4)) = 1/2" } \textcolor{g r e e n}{\sqrt{}}$

Since $\sin \left(a\right) = 3$ is outside the range of the sine function, the only valid solution will be $\sin \left(a\right) = \frac{1}{2}$.

The value of $a$ that corresponds to this value is

$a = \textcolor{g r e e n}{\frac{\pi}{6}} \text{ }$ or $\text{ } a = \textcolor{g r e e n}{{30}^{\circ}}$

Do a quick check to make sure that the calculations are correct

$7 \cdot \tan \left(\frac{\pi}{6}\right) + 2 \cdot \cos \left(\frac{\pi}{6}\right) = 5 \cdot \sec \left(\frac{\pi}{6}\right)$

$7 \cdot \frac{\sqrt{3}}{3} + 2 \cdot \frac{\sqrt{3}}{2} = 5 \cdot \frac{2}{\sqrt{3}}$

$14 \sqrt{3} + 6 \sqrt{3} = 20 \sqrt{3} \text{ } \textcolor{g r e e n}{\sqrt{}}$