Question

#QualityQuestion
@Trigonometry

If sec 4θ − sec 2θ = 2, then the general value of θ is?​

Answer 1

Answer:

π/2 + kπ, 3π/10 + πk, π/10 + πk

-π/2 + kπ, -3π/10 + πk, -π/10 + πk

Step-by-step explanation:let 2theta(A)= x

=> sec2x - secx = 2

=> 1/cos2x - 1/cosx = 2

=> cosx - cos2x = 2cosxcos2x

=> cosx - [2cos²x - 1] = 2cosx * cos2x

=> 2cos²x - cosx - 1 + 2cosx cos2x = 0

=> 2cos²x - cosx - 1 + 2cosx[2cos²x - 1] = 0

=> 2cos²x - cosx - 1 + 4cos³x - 2cosx = 0

Let cosx = t

=> 4t³ + 2t² - 3t - 1 = 0

=> 4t³ + 4t² - 2t² - 2t - t - 1 = 0

=> 4t²(t + 1) - 2t(t + 1) + (1 + t) = 0

=> (t + 1)(4t² - 2t + 1) = 0

=> t = -1, or t = (1 ± √5)/4 [ignoring complex one]

Therefore, cosx = -1 → cos(A/2) = cosπ

=> A = π/2 [principal solution]

=> A = π/2 + kπ [general solution]

*if you have the value of inverse cos of (1 ± √5)/4, which is nearly 108° + 2πk and 36° + 2πk. So, A = (108/2)° + πk and (36/2)° + πk. Hence, A is also 54° (=3π/10) and 18°(=π/10).

A = 3π/10 + πk , A = π/10 + πk

(Since sec is +ve even for -ve values)

A = ± 3π/10 + πk, A = ± π/10 + πk

Answer 2

~Solution :-

• Here's the solution for your trigonometric question.

$\sf{sec 4θ - sec 2θ = 2}$

$\to \sf{cos 2θ - cos 4θ = 2 cos 4θ cos 2θ}$

$\to \sf{ -cos 4θ = cos 6θ}$

$\to \sf{ 2 cos 5θ cosθ = 0}$

$\implies \bf \red{⇒ H = \frac{ [h cot 15o]}{ [cot 15o - 1]} \: or \: \frac{nπ}{5 } + \frac{ π}{10}}$

• Hence, the value of $\sf{\theta}$ will be as given above.

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