Question

Sin2x-sin(3п/2-2x)+1=0

Answer 1

Sin2x+cos2x+1=0
2sinxcosx+cos²x-sin²x+sin²x+cos²x=0
2sinxcosx+2cos²x=0
2cosx(sinx+sin(π/2-x)=0
2cosx*2sinπ/4cos(x-π/4)=0
2*2*√2/2cosxcos(x-π/4)=0
2√2cosxcos(x-π/4)=0
cosx=0⇒x=π/2+πn
cos(x-π/4)=0⇒x-π/4=π/2+πn⇒x=3π/4+πn

Answer 2

To solve the equation sin(2x) - sin(3π/2 - 2x) + 1 = 0, we can apply trigonometric identities and simplify the equation. Let's break it down step by step:

1. Start with the equation: sin(2x) - sin(3π/2 - 2x) + 1 = 0

2. Apply the sum-to-product identity for the second term: sin(a) - sin(b) = 2cos((a+b)/2)sin((a-b)/2) sin(2x) - sin(3π/2 - 2x) = 2cos((2x + 3π/2 - 2x)/2)sin((2x - 3π/2 + 2x)/2) sin(2x) - sin(3π/2 - 2x) = 2cos((3π/2 + x)/2)sin((4x - 3π/2)/2) sin(2x) - sin(3π/2 - 2x) = 2cos((3π/4 + x/2))sin(2x - 3π/4)

3. Substitute the simplified expression back into the equation: 2cos((3π/4 + x/2))sin(2x - 3π/4) + 1 = 0

Now, we have a trigonometric equation that involves both sine and cosine functions. Unfortunately, there is no algebraic method to find the exact solutions for this type of equation. We can use numerical methods or calculators to approximate the solutions.