asked by guest
You asked:
Solve the equation
cos ( x ) sin ( x ) = 0 \cos\left( x \right) \sin\left( x \right) = 0 cos ( x ) sin ( x ) = 0 for the variable
x x x .
MathBot Answer:
The real solutions are: x = { 2 π n | n ∈ Z } ≈ { 6.2831853 n | n ∈ Z } x = { π ( 1 + 2 n ) | n ∈ Z } ≈ { 6.2831853 n + 3.1415927 | n ∈ Z } x = { π ( 1 + 4 n ) 2 | n ∈ Z } ≈ { 6.2831853 n + 1.5707963 | n ∈ Z } x = { π ( 3 + 4 n ) 2 | n ∈ Z } ≈ { 6.2831853 n + 4.712389 | n ∈ Z } \begin{aligned}x &= \left\{2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\pi \left(1 + 2 n\right)\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 3.1415927\; \middle|\; n \in \mathbb{Z}\right\}\\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(3 + 4 n\right)}{2}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 4.712389\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned} x x x x = { 2 πn ∣ n ∈ Z } ≈ { 6.2831853 n ∣ n ∈ Z } = { π ( 1 + 2 n ) ∣ n ∈ Z } ≈ { 6.2831853 n + 3.1415927 ∣ n ∈ Z } = { 2 π ( 1 + 4 n ) n ∈ Z } ≈ { 6.2831853 n + 1.5707963 ∣ n ∈ Z } = { 2 π ( 3 + 4 n ) n ∈ Z } ≈ { 6.2831853 n + 4.712389 ∣ n ∈ Z }
