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受欢迎的三角函数 >

1-sin(2x)=cos(2x)

解答

1 - sin (2 x) = cos (2 x)

解答

x = 2 πn, x = π + 2 πn, x = \frac{π}{4} + πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

1 - sin (2 x) = cos (2 x)

两边减去

cos (2 x)

1 - sin (2 x) - cos (2 x) = 0

使用三角恒等式改写

1 - cos (2 x) - sin (2 x)

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

= 1 - cos (2 x) - 2 sin (x) cos (x)

使用倍角公式:

cos (2 x) = 1 - 2 sin^{2} (x)

= 1 - (1 - 2 sin^{2} (x)) - 2 cos (x) sin (x)

化简

1 - (1 - 2 sin^{2} (x)) - 2 cos (x) sin (x) : 2 sin^{2} (x) - 2 cos (x) sin (x)

1 - (1 - 2 sin^{2} (x)) - 2 cos (x) sin (x)

- (1 - 2 sin^{2} (x)) : - 1 + 2 sin^{2} (x)

- (1 - 2 sin^{2} (x))

打开括号

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

使用加减运算法则

- (- a) = a, - (a) = - a

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

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

1 - 1 = 0

= 2 sin^{2} (x) - 2 cos (x) sin (x)

= 2 sin^{2} (x) - 2 cos (x) sin (x)

2 sin^{2} (x) - 2 cos (x) sin (x) = 0

分解

2 sin^{2} (x) - 2 cos (x) sin (x) : 2 sin (x) (sin (x) - cos (x))

2 sin^{2} (x) - 2 cos (x) sin (x)

使用指数法则:

a^{b + c} = a^{b} a^{c}

sin^{2} (x) = sin (x) sin (x)

= 2 sin (x) sin (x) - 2 sin (x) cos (x)

因式分解出通项

2 sin (x)

= 2 sin (x) (sin (x) - cos (x))

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

分别求解每个部分

sin (x) = 0 or sin (x) - cos (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin (x) - cos (x) = 0 : x = \frac{π}{4} + πn

sin (x) - cos (x) = 0

使用三角恒等式改写

sin (x) - cos (x) = 0

在两边除以

cos (x), cos (x) \neq = 0

\frac{sin ( x ) - cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

化简

\frac{sin ( x )}{cos ( x )} - 1 = 0

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

tan (x) - 1 = 0

tan (x) - 1 = 0

将

1

到右边

tan (x) - 1 = 0

两边加上

1

tan (x) - 1 + 1 = 0 + 1

化简

tan (x) = 1

tan (x) = 1

tan (x) = 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

合并所有解

x = 2 πn, x = π + 2 πn, x = \frac{π}{4} + πn

作图

流行的例子

sin(2x-5 pi/6)=-1/2 ,0<= x<= 2pi

sin (2 x - 5 \frac{π}{6}) = - \frac{1}{2}, 0 \leq x \leq 2 π

sin(x)=1,0<= x<= 2pi

sin (x) = 1, 0 \leq x \leq 2 π

6 csc (θ) - 6 = 0

9sin(x)tan(x)+4tan(x)=0

9 sin (x) tan (x) + 4 tan (x) = 0

solvefor x,f[g]=cos(1/(x^2))

so l v e f or x, f [g] = cos (\frac{1}{x ^{2}})