正切函数不等式/>
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正切函数不等式
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![正切函数不等式/>](/uploads/image/z/17842082-50-2.jpg?t=%E6%AD%A3%E5%88%87%E5%87%BD%E6%95%B0%E4%B8%8D%E7%AD%89%E5%BC%8F%2F%3E)
正切函数是单调增函数.所以,
tan(π/3 - x/2) < √3
-π/2 + mπ ≤ π/3 - x/2 < arctan(√3/3) = π/6 + mπ
-5π/6 + mπ ≤ -x/2 < -π/6 + mπ
5π/3 - 2mπ ≥ x > π/3 - 2mπ
换一下形式,
2kπ + 5π/3≥ x > 2kπ + π/3
其中,k、m 都为任意整数.
再问: 追加悬赏拉,求解另外两题●▽●
再答: 2(sinx)^2 - 5sinx + 2 = (2sinx -1)(sinx -2) > 0所以,sinx > 2 (不成立,因为 sinx ≤ 1)或 sinx < 1/2因此,-π/2 + 2kπ≤ x < π/6 + 2kπ,k 为任意整数sin(2x) - √3cos(2x)= 2*[1/2 *sin(2x) - (√3/2)*cos(2x)]= 2*[cos(π/3)*sin(2x) - sin(π/3)*cos(2x)]=2*sin(2x - π/3) > √2sin(2x - π/3) > √2/2所以,π/4 + 2kπ < 2x - π/3 < 3π/4 + 2kππ/8 + kπ < x - π/6 < 3π/8 + kπ7π/24 + kπ < x < 13π/24 + kπ,k 为任意整数
tan(π/3 - x/2) < √3
-π/2 + mπ ≤ π/3 - x/2 < arctan(√3/3) = π/6 + mπ
-5π/6 + mπ ≤ -x/2 < -π/6 + mπ
5π/3 - 2mπ ≥ x > π/3 - 2mπ
换一下形式,
2kπ + 5π/3≥ x > 2kπ + π/3
其中,k、m 都为任意整数.
再问: 追加悬赏拉,求解另外两题●▽●
再答: 2(sinx)^2 - 5sinx + 2 = (2sinx -1)(sinx -2) > 0所以,sinx > 2 (不成立,因为 sinx ≤ 1)或 sinx < 1/2因此,-π/2 + 2kπ≤ x < π/6 + 2kπ,k 为任意整数sin(2x) - √3cos(2x)= 2*[1/2 *sin(2x) - (√3/2)*cos(2x)]= 2*[cos(π/3)*sin(2x) - sin(π/3)*cos(2x)]=2*sin(2x - π/3) > √2sin(2x - π/3) > √2/2所以,π/4 + 2kπ < 2x - π/3 < 3π/4 + 2kππ/8 + kπ < x - π/6 < 3π/8 + kπ7π/24 + kπ < x < 13π/24 + kπ,k 为任意整数