2sin(x+π/4)sin(x-π/4)=1 x[0;π/2)

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вопрос

To solve the given equation:

2sin(x+π/4)sin(x-π/4) = 1

First, we can use the trigonometric identity:
sin(a+b)sin(a-b) = sin^2(a) - sin^2(b)

So, we have:
2sin(x+π/4)sin(x-π/4) = sin^2(x+π/4) - sin^2(x-π/4)

Now, let's expand out the trigonometric functions:
2[sin(x)cos(π/4) + cos(x)sin(π/4)][sin(x)cos(π/4) - cos(x)sin(π/4)] = sin^2(x + π/4) - sin^2(x - π/4)

2[(sin(x)cos(π/4))^2 - (cos(x)sin(π/4))^2] = sin^2(x + π/4) - sin^2(x - π/4)

2[sin^2(x)cos^2(π/4) - cos^2(x)sin^2(π/4)] = sin^2(x + π/4) - sin^2(x - π/4)

2[sin^2(x)(1/2) - cos^2(x)(1/2)] = sin^2(x + π/4) - sin^2(x - π/4)

sin^2(x)/2 - cos^2(x)/2 = sin^2(x + π/4) - sin^2(x - π/4)

Now, we can simplify this equation further by noting that sin(π/4) = cos(π/4) = 1/√2:

sin^2(x)/2 - cos^2(x)/2 = sin^2(x + π/4) - sin^2(x - π/4)
sin^2(x)/2 - cos^2(x)/2 = sin^2(x)(1/√2)^2 - sin^2(x)(-1/√2)^2
sin^2(x)/2 - cos^2(x)/2 = sin^2(x)/2 - sin^2(x)/2
0 = 0

Since the equation simplifies to 0 = 0, it means that the equation is true for all values of x in the interval [0, π/2). Therefore, the original equation holds for all x in the interval [0, π/2).