asked by guest
You asked:
Solve the equation
3 tan ( x ) 2 − 7 tan ( x ) = − 2 3 {\tan\left( x \right)}^{2} - 7 \tan\left( x \right) = -2 3 tan ( x ) 2 − 7 tan ( x ) = − 2 for the variable
x x x .
MathBot Answer:
The real solutions are: x = { 2 π n + arctan ( 1 3 ) | n ∈ Z } ≈ { 6.2831853 n + 0.32175055 | n ∈ Z } x = { 2 π n + arctan ( 2 ) | n ∈ Z } ≈ { 6.2831853 n + 1.1071487 | n ∈ Z } x = { π + 2 π n + arctan ( 1 3 ) | n ∈ Z } ≈ { 6.2831853 n + 3.4633432 | n ∈ Z } x = { π + 2 π n + arctan ( 2 ) | n ∈ Z } ≈ { 6.2831853 n + 4.2487414 | n ∈ Z } \begin{aligned}x &= \left\{2 \pi n + \arctan{\left(\frac{1}{3} \right)}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 0.32175055\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{2 \pi n + \arctan{\left(2 \right)}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 1.1071487\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\pi + 2 \pi n + \arctan{\left(\frac{1}{3} \right)}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 3.4633432\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\pi + 2 \pi n + \arctan{\left(2 \right)}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 4.2487414\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned} x x x x = { 2 πn + arctan ( 3 1 ) n ∈ Z } ≈ { 6.2831853 n + 0.32175055 ∣ n ∈ Z } = { 2 πn + arctan ( 2 ) ∣ n ∈ Z } ≈ { 6.2831853 n + 1.1071487 ∣ n ∈ Z } = { π + 2 πn + arctan ( 3 1 ) n ∈ Z } ≈ { 6.2831853 n + 3.4633432 ∣ n ∈ Z } = { π + 2 πn + arctan ( 2 ) ∣ n ∈ Z } ≈ { 6.2831853 n + 4.2487414 ∣ n ∈ Z }
