1: 7sin^2x+5sinx-2=0 2: 5sin^2x-21cosx-9=0 3: 5tgx -6ctgx+7=0 4: 4cosx+sinx=0 5: sin^2x-6sinx=0 6: cos6x+cos4x=0 7: sin2x-2sin x=0 8: 3sin2x+2sin^2x=0 9: 7cos2x+18sin^2x-9=0 10: cos2x+11sin x-6=0

Предыдущий
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вопрос

1: The given equation is a quadratic equation in terms of sinx. We can solve it by using the quadratic formula:

sinx = (-5 ± sqrt(5^2 - 47(-2))/2*7
sinx = (-5 ± sqrt(25 + 56))/14
sinx = (-5 ± sqrt(81))/14
sinx = (-5 ± 9)/14

Therefore, sinx = 1 or sinx = -0.5

2: To solve this equation, we can use the Pythagorean identity to replace sin^2x:

5(1-cos^2x)-21cosx-9=0
5-5cos^2x-21cosx-9=0
5cos^2x+21cosx-14=0

Now, we can use the quadratic formula to solve for cosx.

3: This equation involves the trigonometric functions tangent and cotangent. We can rewrite the equation in terms of a single trigonometric function by making the substitution cotgx = 1/tgx:

5tgx - 6(1/tgx) + 7 = 0
5tg^2x - 6 + 7tgx = 0

Now we can solve this quadratic equation for tgx.

4: The given equation involves both cosine and sine functions. We can rewrite it using the Pythagorean identity sin^2x = 1 - cos^2x:

4(1-cos^2x) + sinx = 0
4 - 4cos^2x + sinx = 0
4 - 4(1-sin^2x) + sinx = 0

Solving this equation will give us the value of x.

5: This equation is a quadratic in terms of sinx. We can factor out sinx:

sinx(sinx-6) = 0

Therefore, sinx = 0 or sinx = 6. Since sinx cannot be greater than 1, the only solution is sinx = 0.

6: This equation involves cosine functions. We can rewrite it using the sum-to-product formula for cosine:

cos6x + cos4x = 2cos(5x)cos(x) = 0

From this, we can deduce that either cos(5x) = 0 or cos(x) = 0.

7: We can rewrite the equation using the double angle formula for sine:

2sinxcosx - 2sinx = 0
2sinx(cosx - 1) = 0

Therefore, sinx = 0 or cosx = 1.

8: This equation is a quadratic in terms of sinx. We can factor out sinx:

sinx(3sinx+2) = 0

Therefore, sinx = 0 or sinx = -2/3.

9: To solve this equation, we will use the Pythagorean identity to replace cos2x:

7(1 - 2sin^2x) + 18sin^2x - 9 = 0
7 - 14sin^2x + 18sin^2x - 9 = 0
4sin^2x - 2 = 0

Solving this quadratic equation gives us the value of sinx.

10: We can rewrite the equation using the double angle formula for cosine and the Pythagorean identity:

2cos^2x + 11sinx - 6 = 0
2(1-sin^2x) + 11sinx - 6 = 0

Solving this equation will give us the value of x.