第一题：
sec²x = tanx + 1,0 ≤ x ≤ 2π
1 + tan²x = tanx + 1
tanx(tanx - 1) = 0
tanx = 0 OR tanx = 1
x = 0,π,2π OR x = π/4,π + π/4
x = 0,π/4,π,5π/4,2π
第二题：这题很有挑战性,不像是中学程度的
先说说答案：
x = arccos[√3/4 - (1/4)√(8√3 - 13)] ≈ 78.366°
x = 2π - arccos[√3/4 + (1/4)√(8√3 - 13)] ≈ 311.634°
x = 2π + arccos[√3/4 - (1/4)√(8√3 - 13)] ≈ 438.366°
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3cosx + sin(2x) = 1,0 ≤ x ≤ 10,即0 ≤ x ≤ 572.958°,化为角度比较好看
3cosx + 2sinxcosx = 1
cosx(3 + 2sinx) = 1 ...※
3 + 2sinx = 1/cosx
2sinx = 1/cosx - 3 = (1 - 3cosx)/cosx
sinx = (1 - 3cosx)/(2cosx),代入sin²x + cos²x = 1中
(1 - 3cosx)²/(2cosx)² + cos²x = 1
4cos²x + 1/cos²x + 5 = 6/cosx
4cos⁴x + 5cos²x - 6cosx + 1 = 0,令ρ = cosx
4ρ⁴ + 5ρ² - 6ρ + 1 = 0,这方程用软件计算的
解得ρ = cosx = √3/4 - (1/2)√(2√3 - 13/4)
OR ρ = cosx = √3/4 + (1/2)√(2√3 - 13/4)
参考角x = arccos[√3/4 - (1/4)√(8√3 - 13)] ≈ 78.37°
于是x = 78.37°,281.63°,438.37°
参考角x = arccos[√3/4 + (1/4)√(8√3 - 13)] ≈ 48.37°
于是x = 48.37°,311.63°,408.37°
亦由※部分得：
cosx = 1/(3 + 2sinx),代入sin²x + cos²x = 1中
sin²x + 1/(3 + 2sinx)² = 1
sin²x(3 + 2sinx)² + 1 = (3 + 2sinx)²
sin²x(9 + 4sin²x + 12sinx) + 1 = 9 + 4sin²x + 12sinx
9sin²x + 4sin⁴x + 12sin³x + 1 = 9 + 4sin²x + 12sinx
4sin⁴x + 12sin³x + 5sin²x - 12sinx - 8 = 0
4u⁴ + 12u³ + 5u² - 12u - 8 = 0,这方程用软件计算的
解得u = sinx = - 3/4 + √3/4 - (1/4)√(5 + 4√3)
OR u = sinx = - 3/4 + √3/4 + (1/4)√(5 + 4√3)
参考角x = arcsin[- 3/4 + √3/4 - (1/4)√(5 + 4√3)],这个是虚数,故舍去
参考角x = arcsin[- 3/4 + √3/4 + (1/4)√(5 + 4√3)] ≈ 33.12°
于是x = 33.12°,213.12°,393.12°
将这9个答案逐个代入原方程,只有3个符合
于是x = 78.366°,311.634°,438.666°