## Description

*Directions: Solve each of the following problems.*

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## Explanation & Answer

Here are the solution files.😊

Q1: - 𝑐𝑜𝑡𝑥 − √3 = 0; 𝑥 ∈ [0,2𝜋]

Sol: 𝑐𝑜𝑡𝑥 − √3 = 0

𝑐𝑜𝑡𝑥 = √3

𝑥 = cot −1 (√3)

𝑥=

𝜋

+ 𝑛𝜋

6

As 𝑥 ∈ [0,2𝜋], so, solution is

𝑥=

𝜋

7𝜋

,𝑥 =

6

6

Hence.

Q2: - 0 ≤ 𝑥 ≤ 2𝜋, 𝑆𝑜𝑙𝑣𝑒 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 4𝑠𝑒𝑐𝑥 + 8 = 0.

Sol: 4𝑠𝑒𝑐𝑥 + 8 = 0

4𝑠𝑒𝑐𝑥 = −8

𝑠𝑒𝑐𝑥 = −

8

4

𝑠𝑒𝑐𝑥 = −2

𝑥 = sec −1 (−2)

𝑥=

2𝜋

+ 2𝜋𝑛,

3

𝑥=

4𝜋

+ 2𝜋𝑛

3

As 0 ≤ 𝑥 ≤ 2𝜋, so, solution is

𝑥=

Hence.

Q3: - 2𝑠𝑖𝑛𝑥 = √3; 𝑥 ∈ [0,2𝜋]

Sol: -

2𝜋

4𝜋

,𝑥 =

3

3

2𝑠𝑖𝑛𝑥 = √3

𝑠𝑖𝑛𝑥 =

√3

2

𝑥 = sin−1 (

𝑥=

𝜋

+ 2𝜋𝑛,

3

√3

)

2

𝑥=

2𝜋

+ 2𝜋𝑛

3

As 𝑥 ∈ [0,2𝜋], so, solution is

𝑥=

𝜋

2𝜋

,𝑥 =

3

3

Hence.

Q4: - 0 ≤ 𝑥 ≤ 2𝜋, 𝑆𝑜𝑙𝑣𝑒 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 = 𝑠𝑖𝑛𝑥

Sol: 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 = 𝑠𝑖𝑛𝑥

2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛𝑥 = 0

𝑠𝑖𝑛𝑥(2𝑐𝑜𝑠𝑥 − 1) = 0

𝑠𝑖𝑛𝑥 = 0 𝑜𝑟 2𝑐𝑜𝑠𝑥 − 1 = 0

𝑠𝑖𝑛𝑥 = 0 𝑜𝑟 𝑐𝑜𝑠𝑥 =

1

2

1

𝑥 = sin−1 (0) 𝑜𝑟 𝑥 = cos −1 ( )

2

𝑥 = 0 + 2𝜋𝑛, 𝑥 = 𝜋 + 2𝜋𝑛 𝑜𝑟 𝑥 =

𝜋

5𝜋

+ 2𝜋𝑛, 𝑥 =

+ 2𝜋𝑛

3

3

As 0 ≤ 𝑥 ≤ 2𝜋, so, solution is

𝑥 = 0, �...

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