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Delvig [45]
3 years ago
14

Match the equations with their solutions over the interval [0, 2π].

Mathematics
2 answers:
alexgriva [62]3 years ago
6 0

Answer:

Step-by-step explanation:

A). cos(x)tan(x) = \frac{1}{2}

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

sin(x) = \frac{1}{2}

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

Since sine is positive in 1st and 2nd quadrant.

Therefore, x = \frac{\pi }{6} , \frac{5\pi }{6}

B). sec(x)cot(x) + 2 = 0

[tex]\frac{1}{cos(x)}\frac{cos(x)}{sin(x)}=(-2)[/tex]

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

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

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

Since sine is negative in 3rd and 4th quadrant.

Therefore, x = \frac{7\pi }{6} and \frac{11\pi }{6}

C). sin(x)cot(x) + \frac{1}{\sqrt{2} } = 0

[sin(x)]\frac{cos(x)}{sin(x)}=-\frac{1}{\sqrt{2}}

cosx = -\frac{1}{\sqrt{2}}

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

Since cosine is negative in 2nd and 3rd quadrant.

Therefore, x = \frac{3\pi }{4} and \frac{5\pi }{4}

D). csc(x)tan(x) - 2 = 0

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

\frac{1}{cosx}=2

cos(x) = \frac{1}{2}

Since x is positive in 1st and 4th quadrant.

x = \frac{\pi }{3} and \frac{5\pi }{3}

nignag [31]3 years ago
5 0

I solved this using a scientific calculator and in radians mode since the given x's is between 0 to 2π. After substitution, the correct pairs are:

cos(x)tan(x) – ½ = 0 → π/6 and 5π/6

cos(π/6)tan(π/6) – ½ = 0

cos(5π/6)tan(5π/6) – ½ = 0

 

sec(x)cot(x) + 2 = 0 → 7π/6 and 11π/6

sec(7π/6)cot(7π/6) + 2 = 0

sec(11π/6)cot(11π/6) + 2 = 0

 

sin(x)cot(x) + 1/sqrt2 = 0 → 3π/4 and 5π/4

sin(3π/4)cot(3π/4) + 1/sqrt2 = 0

sin(5π/4)cot(5π/4) + 1/sqrt2 = 0

 

csc(x)tan(x) – 2 = 0 → π/3 and 5π/3

csc(π/3)tan(π/3) – 2 = 0

csc(5π/3)tan(5π/3) – 2 = 0

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