2sin(2x)/1-sin^2(x)=5 can convert to tan(x)=1,25

2sin(2x)/1-sin^2(x)=5 can convert to tan(x)=1,25

a) Show that the equation \(\tfrac{2\sin(2x)} { 1-\sin^2(x) } = 5\) can convert  to tan(x)=1,25

\(\begin{array}{|rcll|} \hline \dfrac{2\sin(2x)} { 1-\sin^2(x) } &=& 5 \quad & | \quad : 2 \\\\ \dfrac{\sin(2x)} { 1-\sin^2(x) } &=& \frac{5}{2} \\\\ \dfrac{\sin(2x)} { 1-\sin^2(x) } &=&2.5 \quad & | \quad \sin(2x) = 2\sin(x)\cos(x) \\\\ \dfrac{2\sin(x)\cos(x)} { 1-\sin^2(x) } &=& 2.5 \quad & | \quad 1-\sin^2(x) = \cos^2(x) \\\\ \dfrac{2\sin(x)\cos(x)} { \cos^2(x) } &=& 2.5 \quad & | \quad : \cos(x) \\\\ \dfrac{2\sin(x)} { \cos(x) } &=& 2.5 \quad & | \quad : 2 \\\\ \dfrac{\sin(x)} { \cos(x) } &=& 1.25 \quad & | \quad \dfrac{\sin(x)} { \cos(x) } = \tan(x) \\\\ \tan(x) &=& 1.25 \\ \hline \end{array}\)

b) Solv the ecvation tan(x)=1,25 fully

\(\begin{array}{|rcll|} \hline \tan(x) &=& 1.25 \quad & | \quad \arctan() \text{ both sides }\\ x &=& \arctan(1.25) + n\cdot \pi \quad & \quad n \in Z \\ x &=& 51.3401917459^{\circ} + n\cdot \pi \quad & \quad n \in Z \\ \hline \end{array}\)

2sin (2x)

__________    =     5

1 - sin^2 (x)

2sin(x)cos(x)   

___________  =    5

cos^2(x)

2sin (x)

______         =     5

cos (x)

2tan(x)  =  5              divide both sides by 2

tan (x)   = 5/2    = 1.25