What is the general solution for sin(2x)=sin^2(2x) ?

The general solution for sin(2x)=sin^2(2x) is x= pi/4+pin,x=pin,x= pi/2+pin

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sin(2x)=sin^2(2x)

sin (2 x) = sin^{2} (2 x)

x = \frac{π}{4} + πn, x = πn, x = \frac{π}{2} + πn

Solution steps

sin (2 x) = sin^{2} (2 x)

Solve by substitution

sin (2 x) = 1, sin (2 x) = 0

sin (2 x) = 1 : x = \frac{π}{4} + πn

sin (2 x) = 0 : x = πn, x = \frac{π}{2} + πn

Combine all the solutions

x = \frac{π}{4} + πn, x = πn, x = \frac{π}{2} + πn

- 8 sin (2 θ + 4 5^{\circ}) = - 4, 0 \leq x < 36 0^{\circ}

sin (b) = \frac{10.21 ( sin ( 61.3 6 ^{\circ} ) )}{11.47}

cos (θ) = 0.513

cot (x) = - \frac{24}{7}

sin (θ) = - \frac{2}{3}, 18 0^{\circ} < θ < 27 0^{\circ}

What is the general solution for sin(2x)=sin^2(2x) ?
The general solution for sin(2x)=sin^2(2x) is x= pi/4+pin,x=pin,x= pi/2+pin