Question

1) tgx+2ctgx=32)7sin^2x+cos^2x=5sinx​

Answer 1

Ответ:

решение на фотографии

Answer 2

Let's solve the two trigonometric equations separately:

1. tg(x) + 2ctg(x) = 3

First, let's use some trigonometric identities: ctg(x) = 1/tan(x)

Now, we can rewrite the equation as: tan(x) + 2/tan(x) = 3

Multiply both sides by tan(x) to eliminate the denominators: tan^2(x) + 2 = 3tan(x)

Now, let's move all terms to one side of the equation: tan^2(x) - 3tan(x) + 2 = 0

This is a quadratic equation in terms of tan(x). Let's factor it: (tan(x) - 1)(tan(x) - 2) = 0

Now, we have two possible solutions:

1. tan(x) - 1 = 0 tan(x) = 1

2. tan(x) - 2 = 0 tan(x) = 2

Now, let's find the solutions for x for each case:

1. tan(x) = 1 x = arctan(1) x = π/4 + πn, where n is an integer

2. tan(x) = 2 x = arctan(2) x = π/4 + πn, where n is an integer

So, the solutions for the first equation are: x = π/4 + πn, where n is an integer

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2. 7sin^2(x) + cos^2(x) = 5sin(x)

First, let's use the trigonometric identity sin^2(x) + cos^2(x) = 1: 7sin^2(x) + 1 - 5sin(x) = 0

Now, let's rearrange the terms: 7sin^2(x) - 5sin(x) + 1 = 0

This is a quadratic equation in terms of sin(x). Let's solve it. You can use the quadratic formula:

sin(x) = [-b ± √(b^2 - 4ac)] / (2a)

In this case, a = 7, b = -5, and c = 1:

sin(x) = [5 ± √((-5)^2 - 4(7)(1))] / (2(7))

sin(x) = [5 ± √(25 - 28)] / 14

sin(x) = [5 ± √(-3)] / 14

Since the square root of -3 is imaginary (i.e., √(-3) = i√3), there are no real solutions to this equation.

Therefore, there are no real solutions for the second equation.