Question

If sin^2 (x) + cos^4 (x) = 13/16, then general solution for x is ?

Answer 1

If sin²x + cos⁴x = 13/16 , then general solution for x is ?

solution : sin²x + cos⁴x = 13/16

⇒1 - cos²x + cos⁴x = 13/16

⇒cos⁴x - cos²x + 1 = 13/16

⇒16cos⁴x - 16cos²x + 16 = 13

⇒16cos⁴x - 16cos²x + 3 = 0

let cos²x = t

⇒16t² - 16t + 3 = 0

⇒t = {16 ± √(16² - 4 × 16 × 3)}/32

= {16 ± 8)/32

= 1/4, 3/4

so, cos²x = 1/4 , 3/4

for cos²x = 1/4

⇒(cosx)² = (1/2)² = (cosπ/3)²

x = nπ ± (π/3)

for cos²x = 3/4

⇒(cosx)² = (√3/2)² = (cosπ/6)²

x = nπ ± (π/6)

[note : cos²x = cos²α then, x = nπ ± α ]

Therefore the general solutions of given trigonometric equation are ; nπ ± (π/3), nπ ± (π/6)

Answer 2

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Given,

sin²x + cos⁴x = 13/16

To Find:-

General solution of "X"

Calculations:-

sin²x + cos⁴x = 13/16

We know that,

sin²x = 1 - cos²x

⇒1 - cos²x + cos⁴x = 13/16

⇒cos⁴x - cos²x + 1 = 13/16

⇒16cos⁴x - 16cos²x + 16 = 13

⇒16cos⁴x - 16cos²x + 3 = 0

Now,

cos²x = y

16y² - 16y+ 3 = 0

=y = {16 ± √(16² - 4 × 16 × 3)}/32

= {16 ± 8)/32

= 1/4, 3/4

so, cos²x = 1/4 , 3/4

for cos²x = 1/4

⇒(cosx)² = (1/2)² = (cosπ/3)²

x = nπ ± (π/3)

for cos²x = 3/4

⇒(cosx)² = (√3/2)² = (cosπ/6)²

x = nπ ± (π/6)