sec4x-sec2x=2
solve for x
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Answered by
35
Sec[2x] = 1/(2Cos²(x) - 1)
Sec[4x] = 1/(2((2Cos²(x) - 1)² - 1)
Let z = Cos²(x)
Then we have:
1/(2(2z - 1)² - 1) - 1/(2z - 1) = 2
which eventually reduces to the quadratic equation:
16z² - 20z + 5 = 0
with solutions z = (1/8)(5 ± √5)
Sec[4x] = 1/(2((2Cos²(x) - 1)² - 1)
Let z = Cos²(x)
Then we have:
1/(2(2z - 1)² - 1) - 1/(2z - 1) = 2
which eventually reduces to the quadratic equation:
16z² - 20z + 5 = 0
with solutions z = (1/8)(5 ± √5)
aswintrichy:
50
Answered by
10
Answer: x=2nπ+(π/2)
and x=(2nπ/5)+(π/10)
Step-by-step explanation:
Use concept of "solution of trigonometric equations"
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