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受欢迎的三角函数 >

9sin^2(x)tan(x)=16tan(x)

解答

9 sin^{2} (x) tan (x) = 16 tan (x)

解答

x = πn

+1

度数

x = 0^{\circ} + 18 0^{\circ} n

求解步骤

9 sin^{2} (x) tan (x) = 16 tan (x)

两边减去

16 tan (x)

9 sin^{2} (x) tan (x) - 16 tan (x) = 0

分解

9 sin^{2} (x) tan (x) - 16 tan (x) : tan (x) (3 sin (x) + 4) (3 sin (x) - 4)

9 sin^{2} (x) tan (x) - 16 tan (x)

因式分解出通项

tan (x)

= tan (x) (9 sin^{2} (x) - 16)

分解

9 sin^{2} (x) - 16 : (3 sin (x) + 4) (3 sin (x) - 4)

9 sin^{2} (x) - 16

将

9 sin^{2} (x) - 16

改写为

(3 sin (x))^{2} - 4^{2}

9 sin^{2} (x) - 16

将

9

改写为

3^{2}

= 3^{2} sin^{2} (x) - 16

将

16

改写为

4^{2}

= 3^{2} sin^{2} (x) - 4^{2}

使用指数法则:

a^{m} b^{m} = (ab)^{m}

3^{2} sin^{2} (x) = (3 sin (x))^{2}

= (3 sin (x))^{2} - 4^{2}

= (3 sin (x))^{2} - 4^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(3 sin (x))^{2} - 4^{2} = (3 sin (x) + 4) (3 sin (x) - 4)

= (3 sin (x) + 4) (3 sin (x) - 4)

= tan (x) (3 sin (x) + 4) (3 sin (x) - 4)

tan (x) (3 sin (x) + 4) (3 sin (x) - 4) = 0

分别求解每个部分

tan (x) = 0 or 3 sin (x) + 4 = 0 or 3 sin (x) - 4 = 0

tan (x) = 0 : x = πn

tan (x) = 0

tan (x) = 0

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 = 0 + πn

x = 0 + πn

解

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

3 sin (x) + 4 = 0 :

无解

3 sin (x) + 4 = 0

将

4

到右边

3 sin (x) + 4 = 0

两边减去

4

3 sin (x) + 4 - 4 = 0 - 4

化简

3 sin (x) = - 4

3 sin (x) = - 4

两边除以

3

3 sin (x) = - 4

两边除以

3

\frac{3 sin ( x )}{3} = \frac{- 4}{3}

化简

sin (x) = - \frac{4}{3}

sin (x) = - \frac{4}{3}

- 1 \leq sin (x) \leq 1

无解

3 sin (x) - 4 = 0 :

无解

3 sin (x) - 4 = 0

将

4

到右边

3 sin (x) - 4 = 0

两边加上

4

3 sin (x) - 4 + 4 = 0 + 4

化简

3 sin (x) = 4

3 sin (x) = 4

两边除以

3

3 sin (x) = 4

两边除以

3

\frac{3 sin ( x )}{3} = \frac{4}{3}

化简

sin (x) = \frac{4}{3}

sin (x) = \frac{4}{3}

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = πn

作图

流行的例子

3sin(2x)=5cos^2(2x)-1

3 sin (2 x) = 5 cos^{2} (2 x) - 1

sin(x)=2,sin^2(x)=0

sin (x) = 2, sin^{2} (x) = 0

cos(θ)=-sqrt(2)

cos (θ) = - 2

-8cos(2x)=0,(-pi,pi)

- 8 cos (2 x) = 0, (- π, π)

1.66 = sin (x)