Directions: Solve each of the following problems.​

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RBcryy

Mathematics

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Directions: Solve each of the following problems.


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1. Cot x - V3 = 0; x [0,2n] 2.0 SXS 2n, Solve the equation 4 Sec X + 8 = 0. 3. 2 Sin x = 73 ; x [0,2n] 4.0 sxs 2n, Solve the equation 2 Sin x Cos x = Sin x. VE 5. Cos x = .; x [0,2n] 6. Csc x Cot x + Cot x = 0; x [0,2n] V3 7.0 SXS 2n, Solve the equation Sin 2x = 2' 8.0 SXS 2n, Solve the equation Sin 20 = -1. V2 9.0 SXS 2n, Solve the equation Cos 20 = 10. 3 tan x = 3; x [O SXS 2n].
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Explanation & Answer

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Q1: - π‘π‘œπ‘‘π‘₯ βˆ’ √3 = 0; π‘₯ ∈ [0,2πœ‹]
Sol: π‘π‘œπ‘‘π‘₯ βˆ’ √3 = 0
π‘π‘œπ‘‘π‘₯ = √3
π‘₯ = cot βˆ’1 (√3)
π‘₯=

πœ‹
+ π‘›πœ‹
6

As π‘₯ ∈ [0,2πœ‹], so, solution is
π‘₯=

πœ‹
7πœ‹
,π‘₯ =
6
6

Hence.
Q2: - 0 ≀ π‘₯ ≀ 2πœ‹, π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 4𝑠𝑒𝑐π‘₯ + 8 = 0.
Sol: 4𝑠𝑒𝑐π‘₯ + 8 = 0
4𝑠𝑒𝑐π‘₯ = βˆ’8
𝑠𝑒𝑐π‘₯ = βˆ’

8
4

𝑠𝑒𝑐π‘₯ = βˆ’2
π‘₯ = sec βˆ’1 (βˆ’2)
π‘₯=

2πœ‹
+ 2πœ‹π‘›,
3

π‘₯=

4πœ‹
+ 2πœ‹π‘›
3

As 0 ≀ π‘₯ ≀ 2πœ‹, so, solution is
π‘₯=
Hence.
Q3: - 2𝑠𝑖𝑛π‘₯ = √3; π‘₯ ∈ [0,2πœ‹]
Sol: -

2πœ‹
4πœ‹
,π‘₯ =
3
3

2𝑠𝑖𝑛π‘₯ = √3
𝑠𝑖𝑛π‘₯ =

√3
2

π‘₯ = sinβˆ’1 (
π‘₯=

πœ‹
+ 2πœ‹π‘›,
3

√3
)
2

π‘₯=

2πœ‹
+ 2πœ‹π‘›
3

As π‘₯ ∈ [0,2πœ‹], so, solution is
π‘₯=

πœ‹
2πœ‹
,π‘₯ =
3
3

Hence.
Q4: - 0 ≀ π‘₯ ≀ 2πœ‹, π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 2𝑠𝑖𝑛π‘₯π‘π‘œπ‘ π‘₯ = 𝑠𝑖𝑛π‘₯
Sol: 2𝑠𝑖𝑛π‘₯π‘π‘œπ‘ π‘₯ = 𝑠𝑖𝑛π‘₯
2𝑠𝑖𝑛π‘₯π‘π‘œπ‘ π‘₯ βˆ’ 𝑠𝑖𝑛π‘₯ = 0
𝑠𝑖𝑛π‘₯(2π‘π‘œπ‘ π‘₯ βˆ’ 1) = 0
𝑠𝑖𝑛π‘₯ = 0 π‘œπ‘Ÿ 2π‘π‘œπ‘ π‘₯ βˆ’ 1 = 0
𝑠𝑖𝑛π‘₯ = 0 π‘œπ‘Ÿ π‘π‘œπ‘ π‘₯ =

1
2

1
π‘₯ = sinβˆ’1 (0) π‘œπ‘Ÿ π‘₯ = cos βˆ’1 ( )
2
π‘₯ = 0 + 2πœ‹π‘›, π‘₯ = πœ‹ + 2πœ‹π‘› π‘œπ‘Ÿ π‘₯ =

πœ‹
5πœ‹
+ 2πœ‹π‘›, π‘₯ =
+ 2πœ‹π‘›
3
3

As 0 ≀ π‘₯ ≀ 2πœ‹, so, solution is
π‘₯ = 0, οΏ½...


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