Question

Find the general solution for:-

Answer 1

Answer:

$\rm \displaystyle x \approx \bigg \{ {59.3}^{ \circ} + \frac{n\pi}{2} , - {14.3}^{ \circ} + \frac{n\pi}{2} \bigg \}$

Step-by-step explanation:

we would like to solve the following trigonometric equation:

$\rm \displaystyle \sin(x) \cos(3x) + \cos(x) \sin(3x) = \tan( {140}^{ \circ} )$

the left hand side can be rewritten using angle sum indentity of sin which is given by:

$\rm \displaystyle \sin( \alpha + \beta ) = \sin( \alpha ) \cos( \beta ) + \cos( \alpha ) \sin( \beta )$

therefore Let

• $\alpha = x$
• $\beta = 3x$

Thus substitute:

$\rm \displaystyle \sin(x + 3x) = \tan( {140}^{ \circ} )$

simplify addition:

$\rm \displaystyle \sin(4x) = \tan( {140}^{ \circ} )$

keep in mind that sin(t)=sin(π-t) saying that there're two equation to solve:

$\begin{cases} \rm \displaystyle \sin(4x) = \tan( {140}^{ \circ} ) \\ \\ \displaystyle \sin(\pi - 4x) = \tan( {140}^{ \circ} ) \end{cases}$

take inverse trig and that yields:

$\begin{cases} \rm \displaystyle 4x= { \sin}^{ - 1} ( \tan( {140}^{ \circ} ) ) \\ \\ \displaystyle \pi - 4x = { \sin}^{ - 1}( \tan( {140}^{ \circ} ) ) \end{cases}$

add π to both sides of the second equation and that yields:

$\begin{cases} \rm \displaystyle 4x= { \sin}^{ - 1} ( \tan( {140}^{ \circ} ) ) \\ \\ \displaystyle - 4x = { \sin}^{ - 1}( \tan( {140}^{ \circ} ) ) + \pi\end{cases}$

sin function has a period of 2nπ thus add the period:

$\begin{cases} \rm \displaystyle 4x= { \sin}^{ - 1} ( \tan( {140}^{ \circ} ) ) + 2n\pi\\ \\ \displaystyle - 4x = { \sin}^{ - 1}( \tan( {140}^{ \circ} ) ) + \pi + 2n\pi\end{cases}$

divide I equation by 4 and II by -4 which yields:

$\begin{cases} \rm \displaystyle x= \frac{ { \sin}^{ - 1} ( \tan( {140}^{ \circ} ) ) }{4} + \frac{n\pi}{2} \\ \\ \displaystyle x = - \frac{{ \sin}^{ - 1}( \tan( {140}^{ \circ} ) ) + \pi}{4} - \frac{n\pi}{2} \end{cases}$

recall that,-½(nπ)=½(nπ) therefore,

$\begin{cases} \rm \displaystyle x= \frac{ { \sin}^{ - 1} ( \tan( {140}^{ \circ} ) ) }{4} + \frac{n\pi}{2} \\ \\ \displaystyle x = - \frac{{ \sin}^{ - 1}( \tan( {140}^{ \circ} ) ) + \pi}{4} + \frac{n\pi}{2} \end{cases}$

by using a calculator we acquire:

$\begin{cases} \rm \displaystyle x \approx - {14.3}^{ \circ} + \frac{n\pi}{2} \\ \\ \displaystyle x \approx {59.3}^{ \circ} + \frac{n\pi}{2} \end{cases}$

hence,

the general solution for: for the trig equation are

$\rm \displaystyle x \approx \bigg \{ {59.3}^{ \circ} + \frac{n\pi}{2} , - {14.3}^{ \circ} + \frac{n\pi}{2} \bigg \}$