Given that Tan x = 8/12. Find 2cos x + sin^2x
You asked:
Given \(\tan\left( x \right) = \frac{8}{12}\) evaluate the expression \(2 \cdot \cos\left( x \right) + {\sin\left( x \right)}^{2}\).
MathBot Answer:
The principal value in \(- \frac{\pi}{2} < x < \frac{\pi}{2}\) is \(x \approx 0.5880026\)
The real solutions are: \[\begin{aligned}x &= \left\{2 \pi n + \arctan{\left(\frac{2}{3} \right)}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 0.5880026\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\pi + 2 \pi n + \arctan{\left(\frac{2}{3} \right)}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 3.7295953\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned}\]
By substituting the principal value into the expression, we find that: $$\begin{aligned}\sin^{2}{\left(x \right)} + 2 \cos{\left(x \right)} \approx 1.9717929\end{aligned}$$