how to solve sin4x + sin 2x=0
asked by guest
You asked:
Solve the equation
sin ( 4 x ) + sin ( 2 x ) = 0 \sin\left( 4 x \right) + \sin\left( 2 x \right) = 0 sin ( 4 x ) + sin ( 2 x ) = 0 for the variable
x x x .
MathBot Answer:
The real solutions are: x = { π n | n ∈ Z } ≈ { 3.1415927 n | n ∈ Z } x = { π ( 1 2 + n ) | n ∈ Z } ≈ { 3.1415927 n + 1.5707963 | n ∈ Z } x = { π ( 1 3 + n ) | n ∈ Z } ≈ { 3.1415927 n + 1.0471976 | n ∈ Z } x = { π ( 2 + 3 n ) 3 | n ∈ Z } ≈ { 3.1415927 n + 2.0943951 | n ∈ Z } \begin{aligned}x &= \left\{\pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{3.1415927 n\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\pi \left(\frac{1}{2} + n\right)\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{3.1415927 n + 1.5707963\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\pi \left(\frac{1}{3} + n\right)\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{3.1415927 n + 1.0471976\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{\pi \left(2 + 3 n\right)}{3}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{3.1415927 n + 2.0943951\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned} x x x x = { πn ∣ n ∈ Z } ≈ { 3.1415927 n ∣ n ∈ Z } = { π ( 2 1 + n ) n ∈ Z } ≈ { 3.1415927 n + 1.5707963 ∣ n ∈ Z } = { π ( 3 1 + n ) n ∈ Z } ≈ { 3.1415927 n + 1.0471976 ∣ n ∈ Z } = { 3 π ( 2 + 3 n ) n ∈ Z } ≈ { 3.1415927 n + 2.0943951 ∣ n ∈ Z }
