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受欢迎的三角函数 >

2cos(θ)-sqrt(3)=0,0<= θ<= 2pi

解答

2 cos (θ) - 3 = 0, 0 \leq θ \leq 2 π

解答

θ = \frac{π}{6}, θ = \frac{11 π}{6}

+1

度数

θ = 3 0^{\circ}, θ = 33 0^{\circ}

求解步骤

2 cos (θ) - 3 = 0, 0 \leq θ \leq 2 π

将

3

到右边

2 cos (θ) - 3 = 0

两边加上

3

2 cos (θ) - 3 + 3 = 0 + 3

化简

2 cos (θ) = 3

2 cos (θ) = 3

两边除以

2

2 cos (θ) = 3

两边除以

2

\frac{2 cos ( θ )}{2} = \frac{3}{2}

化简

cos (θ) = \frac{3}{2}

cos (θ) = \frac{3}{2}

cos (θ) = \frac{3}{2}

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

在

0 \leq θ \leq 2 π

范围内的解

θ = \frac{π}{6}, θ = \frac{11 π}{6}

作图

流行的例子

-4sin(x)=cos^2(x)+1,0<= x<= 2pi

- 4 sin (x) = cos^{2} (x) + 1, 0 \leq x \leq 2 π

tan(2x)cos(2x)+cot(2x)sin(2x)=1

tan (2 x) cos (2 x) + cot (2 x) sin (2 x) = 1

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

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

sin(x)=(1.2)/(sqrt(10))

sin (x) = \frac{1.2}{10}

solvefor t,x=-3cos(pit)

so l v e f or t, x = - 3 cos (π t)