Question

1. sinx 1/2x=1 2. 2cosx+ √2=0 3. tg x/3 =- √3 4. √2sin( x/2+П/4)=1 5. 2cos(4x- П/6)=√3

Answer 1

Ответ:

решение на фотографиях

Answer 2

To solve these trigonometric equations, we'll go through each one step by step.

1. sin(x/2) = 1/2 To solve this equation, we'll first find the angles where sin(x/2) equals 1/2. The reference angle for sin is π/6, and sin is positive in the first and second quadrants.

   So, x/2 = π/6 + 2πn or x/2 = π - π/6 + 2πn, where n is an integer.

   Now, solve for x: x = π/3 + 4πn or x = 5π/3 + 4πn, where n is an integer.

2. 2cos(x) + √2 = 0 To solve this equation, we need to isolate cos(x):

   2cos(x) = -√2 cos(x) = -√2/2

   This is equivalent to cos(x) = -1/√2. The reference angle for cos is π/4, and cos is negative in the second and third quadrants.

   So, x = 3π/4 + 2πn or x = 5π/4 + 2πn, where n is an integer.

3. tan(x/3) = -√3 To solve this equation, we'll first find the angles where tan(x/3) equals -√3. The reference angle for tan is π/3, and tan is negative in the third and fourth quadrants.

   So, x/3 = -π/3 + πn or x/3 = -2π/3 + πn, where n is an integer.

   Now, solve for x: x = -π + 3πn or x = -2π + 3πn, where n is an integer.

4. √2sin(x/2 + π/4) = 1 To solve this equation, divide both sides by √2:

   sin(x/2 + π/4) = 1/√2

   The reference angle for sin is π/4, and sin is positive in the first and second quadrants.

   So, x/2 + π/4 = π/4 + 2πn or x/2 + π/4 = 3π/4 + 2πn, where n is an integer.

   Now, solve for x: x/2 = 2πn or x/2 = 2πn + π/2, where n is an integer.

5. 2cos(4x - π/6) = √3 To solve this equation, divide both sides by 2:

   cos(4x - π/6) = √3/2

   The reference angle for cos is π/6, and cos is positive in the first and fourth quadrants.

   So, 4x - π/6 = π/6 + 2πn or 4x - π/6 = 5π/6 + 2πn, where n is an integer.

   Now, solve for x: 4x = π/3 + 2πn or 4x = 7π/6 + 2πn, where n is an integer.

   x = π/12 + πn/2 or x = 7π/24 + πn/2, where n is an integer.