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受欢迎的三角函数 >

-2sin^2(θ)+10sin(θ)-3=3sin(θ),0<= θ<360

解答

- 2 sin^{2} (θ) + 10 sin (θ) - 3 = 3 sin (θ), 0^{\circ} \leq θ < 36 0^{\circ}

解答

θ = 3 0^{\circ}, θ = 15 0^{\circ}

+1

弧度

θ = \frac{π}{6}, θ = \frac{5 π}{6}

求解步骤

- 2 sin^{2} (θ) + 10 sin (θ) - 3 = 3 sin (θ), 0^{\circ} \leq θ < 36 0^{\circ}

用替代法求解

- 2 sin^{2} (θ) + 10 sin (θ) - 3 = 3 sin (θ)

令

: sin (θ) = u

- 2 u^{2} + 10 u - 3 = 3 u

- 2 u^{2} + 10 u - 3 = 3 u : u = \frac{1}{2}, u = 3

- 2 u^{2} + 10 u - 3 = 3 u

将

3 u

para o lado esquerdo

- 2 u^{2} + 10 u - 3 = 3 u

两边减去

3 u

- 2 u^{2} + 10 u - 3 - 3 u = 3 u - 3 u

化简

- 2 u^{2} + 7 u - 3 = 0

- 2 u^{2} + 7 u - 3 = 0

使用求根公式求解

- 2 u^{2} + 7 u - 3 = 0

二次方程求根公式:

若

a = - 2, b = 7, c = - 3

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 2 ) ( - 3 )}{2 ( - 2 )}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 2 ) ( - 3 )}{2 ( - 2 )}

7^{2} - 4 (- 2) (- 3) = 5

7^{2} - 4 (- 2) (- 3)

使用法则

- (- a) = a

= 7^{2} - 4 \cdot 2 \cdot 3

数字相乘:

4 \cdot 2 \cdot 3 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

数字相减:

49 - 24 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 7 \pm 5}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 7 + 5}{2 ( - 2 )}, u_{2} = \frac{- 7 - 5}{2 ( - 2 )}

u = \frac{- 7 + 5}{2 ( - 2 )} : \frac{1}{2}

\frac{- 7 + 5}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 7 + 5}{- 2 \cdot 2}

数字相加/相减:

- 7 + 5 = - 2

= \frac{- 2}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{4}

约分:

2

= \frac{1}{2}

u = \frac{- 7 - 5}{2 ( - 2 )} : 3

\frac{- 7 - 5}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 7 - 5}{- 2 \cdot 2}

数字相减:

- 7 - 5 = - 12

= \frac{- 12}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 12}{- 4}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

= \frac{12}{4}

数字相除:

\frac{12}{4} = 3

= 3

二次方程组的解是:

u = \frac{1}{2}, u = 3

u = sin (θ)

代回

sin (θ) = \frac{1}{2}, sin (θ) = 3

sin (θ) = \frac{1}{2}, sin (θ) = 3

sin (θ) = \frac{1}{2}, 0 \leq θ < 36 0^{\circ} : θ = 3 0^{\circ}, θ = 15 0^{\circ}

sin (θ) = \frac{1}{2}, 0 \leq θ < 36 0^{\circ}

sin (θ) = \frac{1}{2}

的通解

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

在

0 \leq θ < 36 0^{\circ}

范围内的解

θ = 3 0^{\circ}, θ = 15 0^{\circ}

sin (θ) = 3, 0 \leq θ < 36 0^{\circ} :

无解

sin (θ) = 3, 0 \leq θ < 36 0^{\circ}

- 1 \leq sin (x) \leq 1

无解

合并所有解

θ = 3 0^{\circ}, θ = 15 0^{\circ}

作图

流行的例子

cot^2(θ)+4csc(θ)=-5

cot^{2} (θ) + 4 csc (θ) = - 5

10sin(a)+7sin(a)+1sin(a)+8sin(a)=48

10 sin (a) + 7 sin (a) + 1 sin (a) + 8 sin (a) = 48

2cos^2(θ)+7sin(θ)=5

2 cos^{2} (θ) + 7 sin (θ) = 5

(2sqrt(3))/3 =sec(θ)

\frac{2 3}{3} = sec (θ)

4(sin(x))^2-15sin(x)-4=0

4 (sin (x))^{2} - 15 sin (x) - 4 = 0