三角方程式 cos^2(x) + 2cos(x) + 1 = 0 を [0, 2pi) で解きます。
2cos^2(x) + cos(x) - 1 = 0、x を求めます。
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【関数#1】0<sinx+cosx+tanxの時,0<sin^3x+cos^3x+tan^3xとなることを示せ!!【難易度★★★★☆☆☆☆☆☆】
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三角方程式: sin (2x) = cos (x) を解きます。
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Solve the Trig equation sin x + cos x = 1 over the interval [0, 2pi)
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