# If 8sinx+3cosx=5 then what is cot x ?

Apr 17, 2017

$\cot x = \frac{3}{2} \pm \frac{5}{4} \sqrt{3}$

#### Explanation:

Given:

$8 \sin x + 3 \cos x = 5$

Subtract $3 \cos x$ from both sides to get:

$8 \sin x = 5 - 3 \cos x$

Square both sides to get:

$64 {\sin}^{2} x = 25 - 30 \cos x + 9 {\cos}^{2} x$

Note that squaring will not introduce spurious solutions in this case since $\sin \left(- x\right) = - \sin \left(x\right)$ but $\cos \left(- x\right) = \cos \left(x\right)$. So both signs are possible.

Use ${\cos}^{2} x + {\sin}^{2} x = 1$ to reexpress the left hand side and get:

$64 - 64 {\cos}^{2} x = 25 - 30 \cos x + 9 {\cos}^{2} x$

Add $64 {\cos}^{2} x - 64$ to both sides to get:

$0 = 73 {\cos}^{2} x - 30 \cos x - 39$

Use the quadratic formula to get:

$\cos x = \frac{30 \pm \sqrt{{\left(- 30\right)}^{2} - 4 \left(73\right) \left(- 39\right)}}{2 \cdot 73}$

$\textcolor{w h i t e}{\cos x} = \frac{30 \pm \sqrt{900 + 11388}}{146}$

$\textcolor{w h i t e}{\cos x} = \frac{30 \pm \sqrt{12288}}{146}$

$\textcolor{w h i t e}{\cos x} = \frac{15}{73} \pm \frac{32}{73} \sqrt{3}$

Then from the original equation:

$\sin x = \frac{5 - 3 \cos x}{8}$

So if $\cos x = \frac{15}{73} + \frac{32}{73} \sqrt{3}$ then:

$\sin x = \frac{5 - 3 \left(\frac{15}{73} + \frac{32}{73} \sqrt{3}\right)}{8}$

$\textcolor{w h i t e}{\sin x} = \frac{40}{73} - \frac{12}{73} \sqrt{3}$

resulting in:

$\cot x = \cos \frac{x}{\sin} x$

$\textcolor{w h i t e}{\cot x} = \frac{\frac{15}{73} + \frac{32}{73} \sqrt{3}}{\frac{40}{73} - \frac{12}{73} \sqrt{3}}$

$\textcolor{w h i t e}{\cot x} = \frac{15 + 32 \sqrt{3}}{40 - 12 \sqrt{3}}$

$\textcolor{w h i t e}{\cot x} = \frac{\left(15 + 32 \sqrt{3}\right) \left(10 + 3 \sqrt{3}\right)}{4 \left(10 - 3 \sqrt{3}\right) \left(10 + 3 \sqrt{3}\right)}$

$\textcolor{w h i t e}{\cot x} = \frac{150 + 45 \sqrt{3} + 320 \sqrt{3} + 288}{4 \left(100 - 27\right)}$

$\textcolor{w h i t e}{\cot x} = \frac{438 + 365 \sqrt{3}}{292}$

$\textcolor{w h i t e}{\cot x} = \frac{3}{2} + \frac{5}{4} \sqrt{3}$

Similarly, if $\cos x = \frac{15}{73} - \frac{32}{73} \sqrt{3}$ then:

$\cot x = \frac{3}{2} - \frac{5}{4} \sqrt{3}$

Apr 17, 2017

$\cot x = \frac{1}{4} \left(6 \pm 5 \sqrt{3}\right) , \mathmr{and} , \frac{3}{2} \pm \frac{5}{4} \sqrt{3.}$

#### Explanation:

Given that, $8 \sin x + 3 \cos x = 5. \ldots . . \left(\ast\right) .$

Dividing by $\sin x \ne 0 , \text{ we get, "8+3cotx=5cscx," &, squaring,}$

$64 + 48 \cot x + 9 {\cot}^{2} x = 25 {\csc}^{2} x = 25 \left(1 + {\cot}^{2} x\right)$

$\Rightarrow 16 {\cot}^{2} x - 48 \cot x = 39$

Completing square, $16 {\cot}^{2} x - 48 \cot x + 36 = 39 + 36 = 75$

$\therefore {\left(4 \cot x - 6\right)}^{2} = 75 \Rightarrow 4 \cot x - 6 = \pm 5 \sqrt{3}$

$\text{Therefore, } \cot x = \frac{1}{4} \left(6 \pm 5 \sqrt{3}\right) , \mathmr{and} , \frac{3}{2} \pm \frac{5}{4} \sqrt{3.}$

The above soln. was derived on the assumption that, $\sin x \ne 0.$

If, $\sin x = 0 ,$ then, $\cos x = \pm 1 ,$ and sub.ing these in $\left(\ast\right) ,$

we get, $\pm 3 = 5$, which is impossible. Hence, $\sin x \ne 0$ holds good.

Enjoy Maths.!