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x^2+2*x*sin(x-pi/12)+2*cos(x-pi/12)=(pi-pi/12)^2-2

發布時間:2025/5/22 人文关怀 27 博士
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We can try to simplify the left-hand side of the equation by using trigonometric identities. First, we notice that the expression "2*x*sin(x-pi/12)" involves the product of two trigonometric functions, which can be written as: 2*x*sin(x-pi/12) = x*sin(x-pi/12) + x*sin(x-pi/12) Next, we can use the sum-to-product identity for sine to rewrite each term as a difference of cosines: x*sin(x-pi/12) = x*(sin(x)*cos(pi/12) - cos(x)*sin(pi/12)) = x*sin(x)*cos(pi/12) - x*cos(x)*sin(pi/12) Similarly, for "2*cos(x-pi/12)", we can use the sum-to-product identity for cosine: 2*cos(x-pi/12) = 2*(cos(x)*cos(pi/12) + sin(x)*sin(pi/12)) = 2*cos(x)*cos(pi/12) + 2*sin(x)*sin(pi/12) Substituting these expressions back into the original equation, we get: x^2 + x*sin(x)*cos(pi/12) - x*cos(x)*sin(pi/12) + x*sin(x)*cos(pi/12) - x*cos(x)*sin(pi/12) + 2*cos(x)*cos(pi/12) + 2*sin(x)*sin(pi/12) = (pi-pi/12)^2 - 2 Simplifying each term, we obtain: x^2 + 2*x*sin(x)*cos(pi/12) + 2*cos(x)*cos(pi/12) = (pi-pi/12)^2 - 2 Now we can use the double-angle formula for cosine to express the product of cosine and cosine(pi/12) as a sum of squares: cos(2*pi/12) = cos(pi/6) = sqrt(3)/2 cos(2*a) = cos^2(a) - sin^2(a) cos(pi/12) = cos(2*pi/12)/2 + sqrt(3)/2*sin(2*pi/12) = (sqrt(3) + 1)/4 (cos(x)*cos(pi/12)) = (cos(x)*(sqrt(3) + 1))/4 Substituting this back into the equation, we get: x^2 + 2*x*sin(x)*(sqrt(3) + 1)/4 + (sqrt(3) + 1)/2*(cos(x))^2 = (pi-pi/12)^2 - 2 Now, we need to eliminate the terms with sin(x) and cos(x)^2. We can do this by using the Pythagorean identity: sin^2(a) + cos^2(a) = 1 cos(x)^2 = 1 - sin(x)^2 Substituting this back into the equation, we get: x^2 + 2*x*sin(x)*(sqrt(3) + 1)/4 + (sqrt(3) + 1)/2*(1-sin(x)^2) = (pi-pi/12)^2 - 2 Simplifying and rearranging, we finally obtain: (5*sqrt(3) - 7)*sin(x)^2 - x*(sqrt(3) + 1)*sin(x) + (x^2 - (pi-pi/12)^2 + 2 - (5*sqrt(3) - 7)*(sqrt(3) + 1)/2) = 0 This is a quadratic equation in sin(x), which can be solved using the quadratic formula. The solutions may be complex or transcendental numbers, and may require numerical approximation. Therefore, we leave the final answer in this form.

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