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受欢迎的三角函数 >

cos^2(x)=(sqrt(3))/2

解答

cos^{2} (x) = \frac{3}{2}

解答

x = 0.37473 \dots + 2 πn, x = 2 π - 0.37473 \dots + 2 πn, x = 2.76685 \dots + 2 πn, x = - 2.76685 \dots + 2 πn

+1

度数

x = 21.47070 \dots^{\circ} + 36 0^{\circ} n, x = 338.52929 \dots^{\circ} + 36 0^{\circ} n, x = 158.52929 \dots^{\circ} + 36 0^{\circ} n, x = - 158.52929 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cos^{2} (x) = \frac{3}{2}

用替代法求解

cos^{2} (x) = \frac{3}{2}

令

: cos (x) = u

u^{2} = \frac{3}{2}

u^{2} = \frac{3}{2} : u = \frac{3}{2}, u = - \frac{3}{2}

u^{2} = \frac{3}{2}

对于

x^{2} = f (a)

解为

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

u = \frac{3}{2}, u = - \frac{3}{2}

u = cos (x)

代回

cos (x) = \frac{3}{2}, cos (x) = - \frac{3}{2}

cos (x) = \frac{3}{2}, cos (x) = - \frac{3}{2}

cos (x) = \frac{3}{2} : x = arccos \frac{3}{2} + 2 πn, x = 2 π - arccos \frac{3}{2} + 2 πn

cos (x) = \frac{3}{2}

使用反三角函数性质

cos (x) = \frac{3}{2}

cos (x) = \frac{3}{2}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos \frac{3}{2} + 2 πn, x = 2 π - arccos \frac{3}{2} + 2 πn

x = arccos \frac{3}{2} + 2 πn, x = 2 π - arccos \frac{3}{2} + 2 πn

cos (x) = - \frac{3}{2} : x = arccos - \frac{3}{2} + 2 πn, x = - arccos - \frac{3}{2} + 2 πn

cos (x) = - \frac{3}{2}

使用反三角函数性质

cos (x) = - \frac{3}{2}

cos (x) = - \frac{3}{2}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos - \frac{3}{2} + 2 πn, x = - arccos - \frac{3}{2} + 2 πn

x = arccos - \frac{3}{2} + 2 πn, x = - arccos - \frac{3}{2} + 2 πn

合并所有解

x = arccos \frac{3}{2} + 2 πn, x = 2 π - arccos \frac{3}{2} + 2 πn, x = arccos - \frac{3}{2} + 2 πn, x = - arccos - \frac{3}{2} + 2 πn

以小数形式表示解

x = 0.37473 \dots + 2 πn, x = 2 π - 0.37473 \dots + 2 πn, x = 2.76685 \dots + 2 πn, x = - 2.76685 \dots + 2 πn

作图

流行的例子

sin (\frac{x}{2}) = \frac{3}{5}

16 cos^{2} (x) - 8 = 0

2 sin (4 θ) = 0

solvefor θ,4sin^2(θ/2)+8cos(θ/2)-7=0

so l v e f or θ, 4 sin^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) - 7 = 0

2sin^2(x)-sin(x)=0,0<= x<= 2pi

2 sin^{2} (x) - sin (x) = 0, 0 \leq x \leq 2 π