Trig Equation?

sin x + cot x = 1

sin x + cos x/sin x = 1

sin²x + cos x = sin x

sin²x − sin x = cos x

(sin²x − sin x)² = cos²x

sin⁴x − 2 sin³x + sin²x = 1 − sin²x

sin⁴x − 2 sin³x + 2 sin²x − 1 = 0

(sin x − 1) (sin³x − sin²x + sin x + 1) = 0

sin x − 1 = 0

sin x = 1

x = π/2 + 2πk

sin³x − sin²x + sin x + 1

sin x ≈ −0.54369

x ≈ −sin⁻¹(0.54369) + 2πk (in Q4)

x ≈ π + sin⁻¹(0.54369) + 2πk (in Q3)

Now we must check for any extraneous solutions that might have been introduced by squaring.

When sin x = 1, then cos x = 0 and cot x = 0, so sin x + cot x = 1 + 0 = 1 ----> ok

When sin x ≈ −0.54369, then:

sin x + cot x = 1

sin x + cos x / sin x = 1

−0.54369 + cos x / −0.54369 = 1

cos x / −0.54369 = 1.54369

cos x = −0.83929

So x is in Q3. This eliminates x ≈ −sin⁻¹(0.54369) + 2πk (in Q4)

Solutions:

x = π/2 + 2πk

x ≈ π + sin⁻¹(0.54369) + 2πk

A graphing calculator shows two solutions in the interval [0, 2π].

.. x = π/2

.. x ≈ 3.71641895111

These solutions are repeated in every 2π interval, so the full solution set can be had by adding 2kπ to these, where k is any integer.

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Most graphing calculators will tell you the root of a function to calculator precision. If not, it is usually not difficult to build a Newton's method iteration function to refine the value of the root.

sin(x) + cot(x) = 1

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

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

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

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

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

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

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

cos(x)^4 - 2 * cos(x)^3 - 2 * cos(x)^2 + cos(x)^2 + 2 * cos(x) + 1 = 1 - cos(x)^2

cos(x)^4 - 2 * cos(x)^3 - cos(x)^2 + 2 * cos(x) + 1 = 1 - cos(x)^2

cos(x)^4 - 2 * cos(x)^3 + 2 * cos(x) = 0

cos(x) * (cos(x)^3 - 2 * cos(x)^2 + 2) = 0

cos(x) = 0

x = pi/2 + pi * k

k is an integer

cos(x)^3 - 2 * cos(x)^2 + 2 = 0

Possible rational roots: cos(x) = -2 , -1 , 1 , 2

2^3 - 2 * 2^2 + 2 = 8 - 8 + 2 = 2

1 - 2 + 2 = 1

-1 - 2 + 2 = -1

(-2)^3 - 2 * (-2)^2 + 2 = -8 - 8 + 2 = -14

So, no rational roots. You'd probably benefit from a solver:

https://www.wolframalpha.com/input/?i=cos(x)%5E3+-...