What is the general solution for 2sin(pi/4+x)*sin(pi/4-x)+sin^2(x)=0 ?

The general solution for 2sin(pi/4+x)*sin(pi/4-x)+sin^2(x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin

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2sin(pi/4+x)*sin(pi/4-x)+sin^2(x)=0

2 sin (\frac{π}{4} + x) \cdot sin (\frac{π}{4} - x) + sin^{2} (x) = 0

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

Solution steps

2 sin (\frac{π}{4} + x) sin (\frac{π}{4} - x) + sin^{2} (x) = 0

Rewrite using trig identities

cos^{2} (x) = 0

Apply rule

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

cos (x) = 0

General solutions for

cos (x) = 0

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

(2 sin (x) - 1) (cos (x) + 1) = 0

7 sin^{2} (x) - 6 = - 3 sin^{2} (x)

tan (b) = 4

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

3 - 3 sin (t) = 33 cos (t)

What is the general solution for 2sin(pi/4+x)*sin(pi/4-x)+sin^2(x)=0 ?
The general solution for 2sin(pi/4+x)*sin(pi/4-x)+sin^2(x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin