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受欢迎的三角函数 >

sin^{22}(x)= 1/2

解答

sin^{22} (x) = \frac{1}{2}

解答

x = 1.32108 \dots + 2 πn, x = π - 1.32108 \dots + 2 πn, x = - 1.32108 \dots + 2 πn, x = π + 1.32108 \dots + 2 πn

+1

度数

x = 75.69275 \dots^{\circ} + 36 0^{\circ} n, x = 104.30724 \dots^{\circ} + 36 0^{\circ} n, x = - 75.69275 \dots^{\circ} + 36 0^{\circ} n, x = 255.69275 \dots^{\circ} + 36 0^{\circ} n

求解步骤

sin^{22} (x) = \frac{1}{2}

用替代法求解

sin^{22} (x) = \frac{1}{2}

令

: sin (x) = u

u^{22} = \frac{1}{2}

u^{22} = \frac{1}{2} : u = 22 \frac{1}{2}, u = - 22 \frac{1}{2}

u^{22} = \frac{1}{2}

对于

x^{n} = f (a)

，n 为偶数，解为

x = n f (a), - n f (a)

u = 22 \frac{1}{2}, u = - 22 \frac{1}{2}

u = sin (x)

代回

sin (x) = 22 \frac{1}{2}, sin (x) = - 22 \frac{1}{2}

sin (x) = 22 \frac{1}{2}, sin (x) = - 22 \frac{1}{2}

sin (x) = 22 \frac{1}{2} : x = arcsin (22 \frac{1}{2}) + 2 πn, x = π - arcsin (22 \frac{1}{2}) + 2 πn

sin (x) = 22 \frac{1}{2}

使用反三角函数性质

sin (x) = 22 \frac{1}{2}

sin (x) = 22 \frac{1}{2}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (22 \frac{1}{2}) + 2 πn, x = π - arcsin (22 \frac{1}{2}) + 2 πn

x = arcsin (22 \frac{1}{2}) + 2 πn, x = π - arcsin (22 \frac{1}{2}) + 2 πn

sin (x) = - 22 \frac{1}{2} : x = arcsin (- 22 \frac{1}{2}) + 2 πn, x = π + arcsin (22 \frac{1}{2}) + 2 πn

sin (x) = - 22 \frac{1}{2}

使用反三角函数性质

sin (x) = - 22 \frac{1}{2}

sin (x) = - 22 \frac{1}{2}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- 22 \frac{1}{2}) + 2 πn, x = π + arcsin (22 \frac{1}{2}) + 2 πn

x = arcsin (- 22 \frac{1}{2}) + 2 πn, x = π + arcsin (22 \frac{1}{2}) + 2 πn

合并所有解

x = arcsin (22 \frac{1}{2}) + 2 πn, x = π - arcsin (22 \frac{1}{2}) + 2 πn, x = arcsin (- 22 \frac{1}{2}) + 2 πn, x = π + arcsin (22 \frac{1}{2}) + 2 πn

以小数形式表示解

x = 1.32108 \dots + 2 πn, x = π - 1.32108 \dots + 2 πn, x = - 1.32108 \dots + 2 πn, x = π + 1.32108 \dots + 2 πn

作图

流行的例子

cos^5(x)=cos(x)

cos^{5} (x) = cos (x)

tan(x)=((1.8)/(3.6))

tan (x) = (\frac{1.8}{3.6})

sec (x + 3 0^{\circ}) = 2

solvefor y,arccos(y/2)=5log_{10}(x/5)

so l v e f or y, arccos (\frac{y}{2}) = 5 lo g_{10} (\frac{x}{5})

sin^{22}(x)= 1/4

sin^{22} (x) = \frac{1}{4}