Question

A)2cos²-5sin+1=0 B) sin4scos2x=sin2xcos4x C) cos2x-sinx=0

Answer 1

$1)\; 2cos^2x-5sinx+1=0\\\\2(1-sin^2x)-5sinx+1=0\\\\2sin^2x+5sinx-3=0\\\\(sinx)_1=-3\ \textless \ -1\; \; net\; reshenij,\\\\ (sinx)_2=\frac{1}{2}\; ,\; \; x=(-1)^{m}\frac{\pi}{6}+\pi m\; ,\; m\in Z\; -\; otvet.$

$2)\; sin4x\cdot cos2x=sin2x\cdot cos4x\\\\sin4x\cdot cos2x-sin2x\cdot cos4x=0\\\\sin(4x-2x)=0\\\\2x=\pi n,\; \; x=\frac{\pi n}{2}\; ,\; n\in Z$

$3)\; cos2x-sinx=0\\\\(1-2sin^2x)-sinx=0\\\\2sin^2x+sinx-1=0\\\\(sinx)_1=-1\; ,\; x=(-1)^{n}\cdot (-\frac{\pi}{2})+\pi n=(-1)^{n+1}\frac{\pi}{2}+\pi n,\; n\in Z\\\\(sinx)_2=\frac{1}{2}\; ,\; x=(-1)^{k}\cdot \frac{\pi}{6}+\pi k\; ,\; k\in Z$

Answer 2

A) 2cos² - 5sin + 1 = 0

To solve the equation 2cos² - 5sin + 1 = 0, we can use some trigonometric identities and techniques. Let's break it down step by step:

Step 1: Rearrange the equation to isolate the cosine term: 2cos² - 5sin + 1 = 0 2cos² = 5sin - 1

Step 2: Use the Pythagorean identity sin² + cos² = 1 to express sin in terms of cos: 2cos² = 5(1 - cos²) - 1 2cos² = 5 - 5cos² - 1 2cos² + 5cos² = 5 - 1 7cos² = 4

Step 3: Divide both sides by 7 to solve for cos²: cos² = 4/7

Step 4: Take the square root of both sides to find cos: cos = ±√(4/7)

So, the solutions for cos are cos = ±√(4/7).

Step 5: Substitute the values of cos into the original equation to find the corresponding values of sin: For cos = √(4/7): 2(√(4/7))² - 5sin + 1 = 0 2(4/7) - 5sin + 1 = 0 8/7 - 5sin + 1 = 0 -5sin = -15/7 sin = 3/7

For cos = -√(4/7): 2(-√(4/7))² - 5sin + 1 = 0 2(4/7) - 5sin + 1 = 0 8/7 - 5sin + 1 = 0 -5sin = -15/7 sin = 3/7

So, the solutions for sin are sin = 3/7.

Therefore, the solutions to the equation 2cos² - 5sin + 1 = 0 are: cos = ±√(4/7) and sin = 3/7.

B) sin(4s)cos(2x) = sin(2x)cos(4x)

To solve the equation sin(4s)cos(2x) = sin(2x)cos(4x), we need to simplify and manipulate the trigonometric expressions. Let's break it down step by step:

Step 1: Use the double-angle identities for sine and cosine: 2sin(s)cos(s)cos(2x) = 2sin(x)cos(x)cos(4x)

Step 2: Rearrange and simplify: sin(2s)cos(2x) = sin(2x)cos(4x)

Step 3: Use the double-angle identity for sine: 2sin(s)cos(s)cos(2x) = 2sin(x)cos(x)cos(4x)

Step 4: Divide both sides by 2 to simplify: sin(s)cos(s)cos(2x) = sin(x)cos(x)cos(4x)

Step 5: Use the double-angle identity for cosine: sin(s)cos(s)(2cos²(x) - 1) = sin(x)cos(x)(8cos⁴(x) - 8cos²(x) + 1)

Step 6: Expand and simplify: 2sin(s)cos(s)cos²(x) - sin(s)cos(s) = 8sin(x)cos(x)cos⁴(x) - 8sin(x)cos(x)cos²(x) + sin(x)cos(x)

Step 7: Use the Pythagorean identity sin²(x) + cos²(x) = 1 to simplify: 2sin(s)cos(s)cos²(x) - sin(s)cos(s) = 8sin(x)cos(x)(1 - sin²(x))⁴ - 8sin(x)cos(x)(1 - sin²(x))² + sin(x)cos(x)

Step 8: Distribute and simplify: 2sin(s)cos(s)cos²(x) - sin(s)cos(s) = 8sin(x)cos(x) - 8sin(x)cos(x)sin²(x) - 8sin(x)cos(x) + 8sin(x)cos(x)sin⁴(x) - 8sin(x)cos(x)sin²(x) + sin(x)cos(x)

Step 9: Combine like terms: 2sin(s)cos(s)cos²(x) - sin(s)cos(s) - 8sin(x)cos(x) + 8sin(x)cos(x)sin⁴(x) - 16sin(x)cos(x)sin²(x) + sin(x)cos(x) = 0

At this point, we have a complicated equation involving multiple trigonometric terms. To find specific solutions, we would need more information about the values of s and x or additional constraints on the equation.

C) cos(2x) - sin(x) = 0

To solve the equation cos(2x) - sin(x) = 0, we can use trigonometric identities and techniques. Let's break it down step by step:

Step 1: Use the double-angle identity for cosine: 2cos²(x) - 1 - sin(x) = 0

Step 2: Rearrange and simplify: 2cos²(x) - sin(x) - 1 = 0

Step 3: Use the Pythagorean identity sin²(x) + cos²(x) = 1 to express cos²(x) in terms of sin(x): 2(1 - sin²(x)) - sin(x) - 1 = 0 2 - 2sin²(x) - sin(x) - 1 = 0 -2sin²(x) - sin(x) + 1 = 0

Step 4: Factor the equation, if possible: (-2sin(x) + 1)(sin(x) + 1) = 0

Step 5: Set each factor equal to 0 and solve for sin(x): -2sin(x) + 1 = 0 or sin(x) + 1 = 0

For -2sin(x) + 1 = 0: -2sin(x) = -1 sin(x) = 1/2

For sin(x) + 1 = 0: sin(x) = -1

Step 6: Find the corresponding values of cos(x) using the Pythagorean identity: For sin(x) = 1/2: cos(x) = ±√(1 - sin²(x)) cos(x) = ±√(1 - (1/2)²) cos(x) = ±√(1 - 1/4) cos(x) = ±√(3/4) cos(x) = ±√3/2

For sin(x) = -1: cos(x) = ±√(1 - sin²(x)) cos(x) = ±√(1 - (-1)²) cos(x) = ±√(1 - 1) cos(x) = ±√0 cos(x) = 0

So, the solutions to the equation cos(2x) - sin(x) = 0 are: sin(x) = 1/2 and cos(x) = ±√3/2 sin(x) = -1 and cos(x) = 0