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受欢迎的三角函数 >

4cos^2(θ)=2,0<= θ<= 2pi

解答

4 cos^{2} (θ) = 2, 0 \leq θ \leq 2 π

解答

θ = \frac{π}{4}, θ = \frac{7 π}{4}, θ = \frac{3 π}{4}, θ = \frac{5 π}{4}

+1

度数

θ = 4 5^{\circ}, θ = 31 5^{\circ}, θ = 13 5^{\circ}, θ = 22 5^{\circ}

求解步骤

4 cos^{2} (θ) = 2, 0 \leq θ \leq 2 π

用替代法求解

4 cos^{2} (θ) = 2

令

: cos (θ) = u

4 u^{2} = 2

4 u^{2} = 2 : u = \frac{1}{2}, u = - \frac{1}{2}

4 u^{2} = 2

两边除以

4

4 u^{2} = 2

两边除以

4

\frac{4 u ^{2}}{4} = \frac{2}{4}

化简

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

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

u = cos (θ)

代回

cos (θ) = \frac{1}{2}, cos (θ) = - \frac{1}{2}

cos (θ) = \frac{1}{2}, cos (θ) = - \frac{1}{2}

cos (θ) = \frac{1}{2}, 0 \leq θ \leq 2 π : θ = \frac{π}{4}, θ = \frac{7 π}{4}

cos (θ) = \frac{1}{2}, 0 \leq θ \leq 2 π

使用反三角函数性质

cos (θ) = \frac{1}{2}

cos (θ) = \frac{1}{2}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn

θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn

在

0 \leq θ \leq 2 π

范围内的解

θ = \frac{π}{4}, θ = \frac{7 π}{4}

cos (θ) = - \frac{1}{2}, 0 \leq θ \leq 2 π : θ = \frac{3 π}{4}, θ = \frac{5 π}{4}

cos (θ) = - \frac{1}{2}, 0 \leq θ \leq 2 π

使用反三角函数性质

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{2}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

在

0 \leq θ \leq 2 π

范围内的解

θ = \frac{3 π}{4}, θ = \frac{5 π}{4}

合并所有解

θ = \frac{π}{4}, θ = \frac{7 π}{4}, θ = \frac{3 π}{4}, θ = \frac{5 π}{4}

作图

流行的例子

sin (θ) = \frac{21}{29}

3tan(θ)+sec^2(θ)=5

3 tan (θ) + sec^{2} (θ) = 5

tan (t) = - \frac{3}{4}

sin (x) = - 0.47

sin (x) = - 0.85