2sint-3cost/2cost-3sint=3

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вопрос

To solve this equation, we can first rewrite it in terms of sine and cosine:

(2sin(t) - 3cos(t)) / (2cos(t) - 3sin(t)) = 3

Next, we can multiply both sides by (2cos(t) - 3sin(t)) to get rid of the denominator:

2sin(t) - 3cos(t) = 3(2cos(t) - 3sin(t))

Expanding both sides, we get:

2sin(t) - 3cos(t) = 6cos(t) - 9sin(t)

Rearranging terms, we get:

2sin(t) + 9sin(t) = 6cos(t) + 3cos(t)

11sin(t) = 9cos(t)

Now we can divide both sides by 11sin(t) to isolate cos(t):

cos(t) = 9/11sin(t)

Finally, we can use the Pythagorean identity sin^2(t) + cos^2(t) = 1 to solve for sin(t) and cos(t) simultaneously.