asked by guest
You asked:
Solve the equation
cos ( 4 x ) = sin ( x ) \cos\left( 4 x \right) = \sin\left( x \right) cos ( 4 x ) = sin ( x ) for the variable
x x x .
MathBot Answer:
The real solutions are: x = { π ( 1 + 4 n ) 2 | n ∈ Z } ≈ { 6.2831853 n + 1.5707963 | n ∈ Z } x = { π ( 7 + 12 n ) 6 | n ∈ Z } ≈ { 6.2831853 n + 3.6651914 | n ∈ Z } x = { π ( 11 + 12 n ) 6 | n ∈ Z } ≈ { 6.2831853 n + 5.7595865 | n ∈ Z } x = { 2 π n + arctan ( 10 − 2 2 5 + 5 ) | n ∈ Z } ≈ { 6.2831853 n + 0.31415927 | n ∈ Z } x = { π + 2 π n + arctan ( 2 + 10 2 5 − 5 ) | n ∈ Z } ≈ { 6.2831853 n + 4.0840704 | n ∈ Z } x = { π − arctan ( 10 − 2 2 5 + 5 ) + 2 π n | n ∈ Z } ≈ { 6.2831853 n + 2.8274334 | n ∈ Z } x = { − arctan ( 2 + 10 2 5 − 5 ) + 2 π + 2 π n | n ∈ Z } ≈ { 6.2831853 n + 5.3407075 | n ∈ Z } \begin{aligned}x &= \left\{\frac{\pi \left(1 + 4 n\right)}{2}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 1.5707963\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{\pi \left(7 + 12 n\right)}{6}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 3.6651914\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{\pi \left(11 + 12 n\right)}{6}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 5.7595865\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{2 \pi n + \arctan{\left(\frac{\sqrt{10} - \sqrt{2}}{2 \sqrt{5 + \sqrt{5}}} \right)}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 0.31415927\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\pi + 2 \pi n + \arctan{\left(\frac{\sqrt{2} + \sqrt{10}}{2 \sqrt{5 - \sqrt{5}}} \right)}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 4.0840704\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\pi - \arctan{\left(\frac{\sqrt{10} - \sqrt{2}}{2 \sqrt{5 + \sqrt{5}}} \right)} + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 2.8274334\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{- \arctan{\left(\frac{\sqrt{2} + \sqrt{10}}{2 \sqrt{5 - \sqrt{5}}} \right)} + 2 \pi + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 5.3407075\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned} x x x x x x x = { 2 π ( 1 + 4 n ) n ∈ Z } ≈ { 6.2831853 n + 1.5707963 ∣ n ∈ Z } = { 6 π ( 7 + 12 n ) n ∈ Z } ≈ { 6.2831853 n + 3.6651914 ∣ n ∈ Z } = { 6 π ( 11 + 12 n ) n ∈ Z } ≈ { 6.2831853 n + 5.7595865 ∣ n ∈ Z } = { 2 πn + arctan ( 2 5 + 5 10 − 2 ) n ∈ Z } ≈ { 6.2831853 n + 0.31415927 ∣ n ∈ Z } = { π + 2 πn + arctan ( 2 5 − 5 2 + 10 ) n ∈ Z } ≈ { 6.2831853 n + 4.0840704 ∣ n ∈ Z } = { π − arctan ( 2 5 + 5 10 − 2 ) + 2 πn n ∈ Z } ≈ { 6.2831853 n + 2.8274334 ∣ n ∈ Z } = { − arctan ( 2 5 − 5 2 + 10 ) + 2 π + 2 πn n ∈ Z } ≈ { 6.2831853 n + 5.3407075 ∣ n ∈ Z }
