Question

Sin2x + sin^2x + cos^2x = 1/2

Answer 1

$sin2x + sin^2x + cos^2x = \frac{1}{2} \\sin2x+(sin^2x+cos^2x)= \frac{1}{2} \\sin2x+1= \frac{1}{2} \\sin2x= \frac{1}{2} -1 \\sin2x=- \frac{1}{2} \\2x=- \frac{\pi}{6} +2\pi n \\x_1=- \frac{\pi}{12} +\pi n,\ n \in Z \\2x=- \frac{5\pi}{6} +2\pi n \\x_2=- \frac{5\pi}{12} +\pi n,\ n \in Z$
Ответ:  $x_1=- \frac{\pi}{12} +\pi n,\ n \in Z ;\ x_2=- \frac{5\pi}{12} +\pi n,\ n \in Z$

Answer 2

To solve the trigonometric equation sin(2x) + sin^2(x) + cos^2(x) = 1/2, we'll need to use trigonometric identities to simplify the expression.

First, let's apply the double-angle identity for sine:

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

Now, the equation becomes:

2 * sin(x) * cos(x) + sin^2(x) + cos^2(x) = 1/2

Next, recall the Pythagorean identity for sine and cosine:

sin^2(x) + cos^2(x) = 1

We can substitute this into the equation:

2 * sin(x) * cos(x) + 1 = 1/2

Now, let's isolate the term with the product of sine and cosine:

2 * sin(x) * cos(x) = 1/2 - 1

2 * sin(x) * cos(x) = -1/2

Now, we can use another identity for the product of sine and cosine:

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

Therefore:

sin(2x) = -1/2

Now, we need to find the solutions for x. The sine function has a value of -1/2 at two specific angles in its domain: -30 degrees (-π/6 radians) and 210 degrees (7π/6 radians) (considering the unit circle).

So, we have:

2x = -30 degrees or 2x = 210 degrees

Now, we'll solve for x:

1. For 2x = -30 degrees:

x = -30 degrees / 2 = -15 degrees

2. For 2x = 210 degrees:

x = 210 degrees / 2 = 105 degrees

So, the solutions for the equation sin(2x) + sin^2(x) + cos^2(x) = 1/2 are:

x = -15 degrees and x = 105 degrees (in degrees)

or

x = -π/12 and x = 7π/12 (in radians)