Given f (1 + SiNx) = 2 + SiNx + cos2x, find f (x) Let u = 1 + SiNx, then SiNx = U-1 (0 ≤ u ≤ 2), then f (U) = - U2 + 3U + 1 (0 ≤ u ≤ 2) So f (x) = - x2 + 3x + 1 (0 ≤ u ≤ 2) How does [f (U) = - U2 + 3U + 1] come from? Ask for advice!

Given f (1 + SiNx) = 2 + SiNx + cos2x, find f (x) Let u = 1 + SiNx, then SiNx = U-1 (0 ≤ u ≤ 2), then f (U) = - U2 + 3U + 1 (0 ≤ u ≤ 2) So f (x) = - x2 + 3x + 1 (0 ≤ u ≤ 2) How does [f (U) = - U2 + 3U + 1] come from? Ask for advice!

Let u = 1 + SiNx, then SiNx = U-1 (0 ≤ u ≤ 2), cos2x = 1-2sin & # 178; X = 1-2 (U-1) &# 178; = 1-2 (U & # 178; - 2U + 1) = 1-2u & # 178; + 4u-2 = - 2U & # 178; + 4u-1, then f (U) = 2 + (U-1) + (- 2U & # 178; + 4u-1) = - 2U & # 178; + 5u is wrong, f (1 + SiNx) = 2 + SiNx +

Let a = (1, the square of SiNx), B = (2, sin2x), where x belongs to (0, pie). If | a * B | = | a | * | B |, then TaNx =?

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Given TaNx = 2, what is the square + 1 of SiNx equal to?

sin²x+1
=(sin²x+sin²x+cos²x)/(sin²x+cos²x)
=(2Sin & # 178; X + cos & # 178; x) / (Sin & # 178; X + cos & # 178; x) the numerator denominator is divided by cos & # 178; X at the same time
=(2tan²x+1)/(tan²x+1)
=(2×2²+1)/(2²+1)
=9/5