Solve 5sinθ+2cosθ=5?

Jun 23, 2018

5sinθ+2cosθ=5

=>2cosθ-5(1-sintheta)=0

=>2(cos^2(θ/2)-sin^2(theta/2))-5(cos^2(theta/2)+sin^2(theta/2)-2cos(theta/2)sin(theta/2))=0

=>2(cos^2(θ/2)-sin^2(theta/2))-5(cos(theta/2)-sin(theta/2))^2=0

$\implies \left(\cos \left(\frac{\theta}{2}\right) - \sin \left(\frac{\theta}{2}\right)\right) \left(2 \cos \left(\frac{\theta}{2}\right) + 2 \sin \left(\frac{\theta}{2}\right) - 5 \cos \left(\frac{\theta}{2}\right) + 5 \sin \left(\frac{\theta}{2}\right)\right) = 0$

$\implies \left(\cos \left(\frac{\theta}{2}\right) - \sin \left(\frac{\theta}{2}\right)\right) \left(7 \sin \left(\frac{\theta}{2}\right) - 3 \cos \left(\frac{\theta}{2}\right)\right) = 0$

When

$\left(\cos \left(\frac{\theta}{2}\right) - \sin \left(\frac{\theta}{2}\right)\right) = 0$

$\implies \tan \left(\frac{\theta}{2}\right) = 1 = \tan \left(\frac{\pi}{4}\right)$

$\implies \frac{\theta}{2} = n \pi + \frac{\pi}{4} \text{ where } n \in \mathbb{Z}$

$\implies \theta = 2 n \pi + \frac{\pi}{2} \text{ where } n \in \mathbb{Z}$

When

$7 \sin \left(\frac{\theta}{2}\right) - 3 \cos \left(\frac{\theta}{2}\right) = 0$

$\implies \tan \left(\frac{\theta}{2}\right) = \frac{3}{7} = \tan \alpha$ (say)

$\implies \frac{\theta}{2} = k \pi + \alpha \text{ where } k \in \mathbb{Z}$

$\implies \theta = 2 k \pi + 2 \alpha \text{ where } k \in \mathbb{Z}$ and $\alpha = {\tan}^{-} 1 \left(\frac{3}{7}\right)$

Jun 23, 2018

$t = {46}^{\circ} 63 + k {360}^{\circ}$
$t = {90}^{\circ} + k {360}^{\circ}$

Explanation:

5sin t + 2cos t = 5
Divide both sides by 5:
$\sin t + \left(\frac{2}{5}\right) \cos t = 1$ (1)
Call $\tan a = \sin \frac{a}{\cos a} = \frac{2}{5}$ --> $a = {21}^{\circ} 80$, and cos a = 0.93.
The equation (1) becomes;
$\sin t \cos a + \sin a . \cos t = \cos a$
$\sin \left(t + {21}^{\circ} 80\right) = 0.93$
Calculator, unit circle and property of sin function give 2 solutions for (t + 21.80):
$\left(t + 21.80\right) = {68.43}^{\circ}$, and
$\left(t + 21.80\right) = 180 - 68.43 = {111}^{\circ}$57#.

a. t + 21.80 = 68.43 +k360
$t = {46}^{\circ} 63 + k {360}^{\circ}$
b. t + 21.80 = 111.57 +k360
$t = {89}^{\circ} 77 + k {360}^{\circ}$, or --> $t = {90}^{\circ} + k {360}^{\circ}$
Check by calculator.
t = 46.63 --> 5sin t = 5(0.726) = 3.63 --> 2cos t = 2(0.686) = 1.37
5sin t + 2cos t = 3.63 + 1.37 = 5. Proved
t = 90^@ --> 5sin t = 5(1) = 5 -> cos t = 0.
5sin t + 2cos t = 5 + 0 = 5. Proved

Jun 23, 2018

$x \in \left\{k \pi + {\left(- 1\right)}^{k} \frac{\pi}{2}\right\} \cup \left\{k \pi + {\left(- 1\right)}^{k} \arcsin \left(\frac{21}{29}\right)\right\} , k \in \mathbb{Z}$.

Explanation:

Given that, $5 \sin \theta + 2 \cos \theta = 5$.

$\therefore 2 \cos \theta = 5 \left(1 - \sin \theta\right)$.

$\text{Squaring, } 4 {\cos}^{2} \theta = 25 {\left(1 - \sin \theta\right)}^{2}$,

$i . e . , 4 \left(1 - {\sin}^{2} \theta\right) = 25 {\left(1 - \sin \theta\right)}^{2}$

$\text{Transposing, } 4 \left(1 - \sin \theta\right) \left(1 + \sin \theta\right) - 25 {\left(1 - \sin \theta\right)}^{2} = 0$.

$\therefore \left(1 - \sin \theta\right) \left\{4 \left(1 + \sin \theta\right) - 25 \left(1 - \sin \theta\right)\right\} = 0$,

$\mathmr{and} , \left(1 - \sin \theta\right) \left(29 \sin \theta - 21\right) = 0$.

$\therefore \sin \theta = 1 , \mathmr{and} , \sin \theta = \frac{21}{29.}$

$\sin \theta = 1 = \sin \left(\frac{\pi}{2}\right) \Rightarrow \theta = k \pi + {\left(- 1\right)}^{k} \frac{\pi}{2} , k \in \mathbb{Z}$.

$\sin \theta = \frac{21}{29} = \sin \left(\arcsin \left(\frac{21}{29}\right)\right) , \text{ gives, }$

$\theta = k \pi + {\left(- 1\right)}^{k} \arcsin \left(\frac{21}{29}\right) , k \in \mathbb{Z}$.

Altogether, we get ,

$x \in \left\{k \pi + {\left(- 1\right)}^{k} \frac{\pi}{2}\right\} \cup \left\{k \pi + {\left(- 1\right)}^{k} \arcsin \left(\frac{21}{29}\right)\right\} , k \in \mathbb{Z}$.