19.62sec^2(x) - 40tan(x) - 0.8 = 0
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You asked:
Solve the equation
19.62 sec ( x ) 2 − 40 tan ( x ) − 0.8 = 0 19.62 {\sec\left( x \right)}^{2} - 40 \tan\left( x \right) - 0.8 = 0 19.62 sec ( x ) 2 − 40 tan ( x ) − 0.8 = 0 for the variable
x x x .
MathBot Answer:
The real solutions are: x = { − arctan ( 20000 76479 − 961 76879 76479 ) 2 + 2 π n | n ∈ Z } ≈ { 6.2831853 n + 0.63495225 | n ∈ Z } x = { π − arctan ( 20000 76479 − 961 76879 76479 ) 2 + 2 π n | n ∈ Z } ≈ { 6.2831853 n + 3.7765449 | n ∈ Z } x = { π 2 − arctan ( 20000 76479 + 961 76879 76479 ) 2 + 2 π n | n ∈ Z } ≈ { 6.2831853 n + 0.91584674 | n ∈ Z } x = { − arctan ( 20000 76479 + 961 76879 76479 ) 2 + 3 π 2 + 2 π n | n ∈ Z } ≈ { 6.2831853 n + 4.0574394 | n ∈ Z } \begin{aligned}x &= \left\{- \frac{\arctan{\left(\frac{20000}{76479} - \frac{961 \sqrt{76879}}{76479} \right)}}{2} + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 0.63495225\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\pi - \frac{\arctan{\left(\frac{20000}{76479} - \frac{961 \sqrt{76879}}{76479} \right)}}{2} + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 3.7765449\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{\pi}{2} - \frac{\arctan{\left(\frac{20000}{76479} + \frac{961 \sqrt{76879}}{76479} \right)}}{2} + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 0.91584674\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{- \frac{\arctan{\left(\frac{20000}{76479} + \frac{961 \sqrt{76879}}{76479} \right)}}{2} + \frac{3 \pi}{2} + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 4.0574394\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned} x x x x = ⎩ ⎨ ⎧ − 2 arctan ( 76479 20000 − 76479 961 76879 ) + 2 πn n ∈ Z ⎭ ⎬ ⎫ ≈ { 6.2831853 n + 0.63495225 ∣ n ∈ Z } = ⎩ ⎨ ⎧ π − 2 arctan ( 76479 20000 − 76479 961 76879 ) + 2 πn n ∈ Z ⎭ ⎬ ⎫ ≈ { 6.2831853 n + 3.7765449 ∣ n ∈ Z } = ⎩ ⎨ ⎧ 2 π − 2 arctan ( 76479 20000 + 76479 961 76879 ) + 2 πn n ∈ Z ⎭ ⎬ ⎫ ≈ { 6.2831853 n + 0.91584674 ∣ n ∈ Z } = ⎩ ⎨ ⎧ − 2 arctan ( 76479 20000 + 76479 961 76879 ) + 2 3 π + 2 πn n ∈ Z ⎭ ⎬ ⎫ ≈ { 6.2831853 n + 4.0574394 ∣ n ∈ Z }
