Question

Can you please determine if the following answers are true or false and if false justify why it's not true and find the correct answer(s)? If true, justify why they are correct.The solutions to 3sinx + 4=5cosx ; [0,2pi) are 4.813 and 1.069.

Answer 1

$3sinx+4=5cosx$

Square bothsides

$\begin{gathered} (3sinx+4)^2=(5cosx)^2 \\ 9\sin^2x+24sinx+16=25\cos^2x \\ 9\sin^2x+24sinx+16=25(1-\sin^2x) \\ 9\sin^2x+24sinx+16=25-25\sin^2x \\ 9\sin^2x+25\sin^2x+24sinx+16-25=0 \\ 34\sin^2x+24sinx-9=0 \end{gathered}$

On solving the quadratic , we have

$sinx=0.2710,\text{ }\sin x=-0.9769$

From sinx = 0.3710 , we are able to obtain angle x in radian as ;

$\begin{gathered} x=\sin^{-1}(0.2710) \\ x=0.2744rad \end{gathered}$

Sine is also positive in the second quadrant , so we have ;

$\begin{gathered} x=\pi-0.2744 \\ x=2.8672rad \end{gathered}$

Also,

For n=1 ,

x = 4.93 rad

So we can conclude that the answers provided in the question are false .