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受欢迎的三角函数 >

7(sin(x))(cos(x))+cos^2(x)=1

解答

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

解答

x = 2 πn, x = π + 2 πn, x = 1.42889 \dots + πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 81.86989 \dots^{\circ} + 18 0^{\circ} n

求解步骤

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

两边减去

1

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

使用三角恒等式改写

- 1 + cos^{2} (x) + 7 cos (x) sin (x)

使用毕达哥拉斯恒等式:

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

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

= 7 cos (x) sin (x) - sin^{2} (x)

- sin^{2} (x) + 7 cos (x) sin (x) = 0

分解

- sin^{2} (x) + 7 cos (x) sin (x) : sin (x) (- sin (x) + 7 cos (x))

- sin^{2} (x) + 7 cos (x) sin (x)

使用指数法则:

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

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

= - sin (x) sin (x) + 7 sin (x) cos (x)

因式分解出通项

sin (x)

= sin (x) (- sin (x) + 7 cos (x))

sin (x) (- sin (x) + 7 cos (x)) = 0

分别求解每个部分

sin (x) = 0 or - sin (x) + 7 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) + 7 cos (x) = 0 : x = arctan (7) + πn

- sin (x) + 7 cos (x) = 0

使用三角恒等式改写

- sin (x) + 7 cos (x) = 0

在两边除以

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

\frac{- sin ( x ) + 7 cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

化简

- \frac{sin ( x )}{cos ( x )} + 7 = 0

使用基本三角恒等式:

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

- tan (x) + 7 = 0

- tan (x) + 7 = 0

将

7

到右边

- tan (x) + 7 = 0

两边减去

7

- tan (x) + 7 - 7 = 0 - 7

化简

- tan (x) = - 7

- tan (x) = - 7

两边除以

- 1

- tan (x) = - 7

两边除以

- 1

\frac{- tan ( x )}{- 1} = \frac{- 7}{- 1}

化简

tan (x) = 7

tan (x) = 7

使用反三角函数性质

tan (x) = 7

tan (x) = 7

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (7) + πn

x = arctan (7) + πn

合并所有解

x = 2 πn, x = π + 2 πn, x = arctan (7) + πn

以小数形式表示解

x = 2 πn, x = π + 2 πn, x = 1.42889 \dots + πn

作图

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