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2cos(θ)+sqrt(2)=0,0<= θ<= 2pi

解答

2 cos (θ) + 2 = 0, 0 \leq θ \leq 2 π

解答

θ = \frac{3 π}{4}, θ = \frac{5 π}{4}

+1

度数

θ = 13 5^{\circ}, θ = 22 5^{\circ}

求解步骤

2 cos (θ) + 2 = 0, 0 \leq θ \leq 2 π

将

2

到右边

2 cos (θ) + 2 = 0

两边减去

2

2 cos (θ) + 2 - 2 = 0 - 2

化简

2 cos (θ) = - 2

2 cos (θ) = - 2

两边除以

2

2 cos (θ) = - 2

两边除以

2

\frac{2 cos ( θ )}{2} = \frac{- 2}{2}

化简

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{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{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

在

0 \leq θ \leq 2 π

范围内的解

θ = \frac{3 π}{4}, θ = \frac{5 π}{4}

流行的例子

sqrt(3)tan(θ/2)-1=0

3 tan (\frac{θ}{2}) - 1 = 0

tan(x/2)=sqrt(3),0<= x<2pi

tan (\frac{x}{2}) = 3, 0 \leq x < 2 π

sin(x)= 4/(cos(x))

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

tan^3(x)-2tan(x)=tan(x)

tan^{3} (x) - 2 tan (x) = tan (x)

sin^2(x)+2cos^2(x)=0

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