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

The general solution for sin^2(2x)=2sin^2(x)cos^2(x) is x= pi/2+2pin,x=(3pi)/2+2pin,x=2pin,x=pi+2pin

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

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

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 2 πn, x = π + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

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

from both sides

sin^{2} (2 x) - 2 sin^{2} (x) cos^{2} (x) = 0

Rewrite using trig identities

sin^{2} (2 x) - 2 cos^{2} (x) sin^{2} (x)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

Simplify

(2 sin (x) cos (x))^{2} - 2 cos^{2} (x) sin^{2} (x) : 2 sin^{2} (x) cos^{2} (x)

(2 sin (x) cos (x))^{2} - 2 cos^{2} (x) sin^{2} (x)

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

2^{2} = 4

= 4 sin^{2} (x) cos^{2} (x) - 2 cos^{2} (x) sin^{2} (x)

Add similar elements:

4 sin^{2} (x) cos^{2} (x) - 2 cos^{2} (x) sin^{2} (x) = 2 sin^{2} (x) cos^{2} (x)

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

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

2 cos^{2} (x) sin^{2} (x) = 0

Solving each part separately

cos^{2} (x) = 0 or sin^{2} (x) = 0

cos^{2} (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

cos (x) = 0

General solutions for

cos (x) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

sin^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (x) = 0

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 2 πn, x = π + 2 πn

tan (θ) = - \frac{1}{5}

\frac{sin ( 7 0 ^{\circ} )}{25} = \frac{sin ( c )}{26}

\frac{csc ( x )}{2} = 3

cos (2 x) = sin (x) - 2

1.25 sin (3 0^{\circ}) = 1.5 sin (x)

What is the general solution for sin^2(2x)=2sin^2(x)cos^2(x) ?
The general solution for sin^2(2x)=2sin^2(x)cos^2(x) is x= pi/2+2pin,x=(3pi)/2+2pin,x=2pin,x=pi+2pin