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# Trigonometry 10 sin^2(x) + 10 Sin (x) Cos (x) - Cos^2 (x) = 2

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10 sin^2(x) + 10 Sin (x) Cos (x) - Cos^2 (x) = 2. Solve for all values of  X between 0 and 360 degrees

Jun 18, 2018

#1
+22343
+3

10 sin^2(x) + 10 sin (x) cos (x) - cos^2 (x) = 2.

Solve for all values of  x between 0 and 360 degrees

$$\begin{array}{|rcll|} \hline 10 \sin^2(x) + 10 \sin (x) \cos (x) - \cos^2 (x) &=& 2 \\ && \small{\boxed{\sin(x) = \tan(x)\cos(x)}} \\ 10 [\tan(x)\cos(x)]^2 + 10 \tan(x)\cos(x)\cos (x) - \cos^2 (x) &=& 2 \\ 10 \tan^2(x)\cos^2(x) + 10 \tan(x)\cos^2(x) - \cos^2 (x) &=& 2 \\ \cos^2(x) \Big( 10 \tan^2(x) + 10 \tan(x) - 1 \Big) &=& 2 \\ 10 \tan^2(x) + 10 \tan(x) - 1 &=& 2 \cdot \dfrac{1}{\cos^2(x) } \\ && \small{\boxed{ \dfrac{1}{\cos^2(x) } = 1+\tan^2(x)} } \\ 10 \tan^2(x) + 10 \tan(x) - 1 &=& 2 \cdot( 1+\tan^2(x)) \\ 10 \tan^2(x) + 10 \tan(x) - 1 &=& 2 +2\tan^2(x) \\ 8 \tan^2(x) + 10 \tan(x) - 3 &=& 0 \\ && \large{\boxed{\tan(x) = z}} \\ 8 z^2 + 10z - 3 &=& 0 \\\\ z&=& \frac{-10\pm \sqrt{100-4\cdot 8\cdot(-3)} } {2\cdot 8} \\ z&=& \frac{-10\pm \sqrt{196} } {16} \\ z&=& \frac{-10\pm 14 } {16} \\\\ z_1 &=& \dfrac{-10 + 14 } {16} \\ \mathbf{z_1} &\mathbf{=}& \mathbf{\dfrac{1} {4}} \\\\ z_2 &=& \dfrac{-10 - 14 } {16} \\ \mathbf{z_2} &\mathbf{=}& -\mathbf{\dfrac{3} {2}} \\ \hline \end{array}$$

solutions:

$$\begin{array}{|rcll|} \hline \tan(x) &=& z_1 \\ \tan(x) &=& \frac{1}{4} \\ \mathbf{x} &\mathbf{=}& \mathbf{\arctan(\frac{1}{4}) + n\cdot 180^{\circ} } \quad & | \quad n \in Z \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \tan(x) &=& z_2 \\ \tan(x) &=& -\frac{3}{2} \\ x &=& \arctan(-\frac{3}{2}) + n\cdot 180^{\circ} \\ \mathbf{x} &\mathbf{=}& \mathbf{-\arctan(\frac{3}{2}) + n\cdot 180^{\circ} } \quad & | \quad n \in Z \\ \hline \end{array}$$

x between 0 and $$\mathbf{360^{\circ}}$$

$$\begin{array}{|rcll|} \hline x_1 &=& \arctan(\frac{1}{4}) \\ \mathbf{x_1} &\mathbf{=}& \mathbf{14.0362434679^{\circ}} \\\\ x_2 &=& -\arctan(\frac{3}{2}) + 180^{\circ} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{123.690067526^{\circ}} \\\\ x_3 &=& \arctan(\frac{1}{4})+ 180^{\circ} \\ \mathbf{x_3} &\mathbf{=}& \mathbf{194.036243468^{\circ}} \\\\ x_4 &=& -\arctan(\frac{3}{2}) + 2\cdot 180^{\circ} \\ \mathbf{x_4} &\mathbf{=}& \mathbf{303.690067526^{\circ}} \\ \hline \end{array}$$

Jun 18, 2018
edited by heureka  Jun 18, 2018
#2
+18329
+3

Here is a graphical solution (a bit easier in this case !)

Jun 18, 2018