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受欢迎的三角函数 >

2sin(x)-sqrt(3)=0,0<= x<= 2pi

解答

2 sin (x) - 3 = 0, 0 \leq x \leq 2 π

解答

x = \frac{π}{3}, x = \frac{2 π}{3}

+1

度数

x = 6 0^{\circ}, x = 12 0^{\circ}

求解步骤

2 sin (x) - 3 = 0, 0 \leq x \leq 2 π

将

3

到右边

2 sin (x) - 3 = 0

两边加上

3

2 sin (x) - 3 + 3 = 0 + 3

化简

2 sin (x) = 3

2 sin (x) = 3

两边除以

2

2 sin (x) = 3

两边除以

2

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

化简

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

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

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

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

在

0 \leq x \leq 2 π

范围内的解

x = \frac{π}{3}, x = \frac{2 π}{3}

作图

流行的例子

8cos(x)+4sin(2x)=0

8 cos (x) + 4 sin (2 x) = 0

1+sin(x)-2sin^2(x)=0

1 + sin (x) - 2 sin^{2} (x) = 0

0 = 2 sin (2 x)

solvefor x,cos(x)-sin(2x)=0

so l v e f or x, cos (x) - sin (2 x) = 0

-2/(sqrt(3))cos(3θ)=1,0<= θ<= 2pi

- \frac{2}{3} cos (3 θ) = 1, 0 \leq θ \leq 2 π