Question

solve this question please:

Answer 1

ya agreed

Answer 2

id throw it in wolfram

Answer 3

not so much stumped as i am lazy

Answer 4

use sin^2 x= 1-cos^2 x

Answer 5

i did that but i can't simplify further

Answer 6

then 16^(1-cos^2 x) = 16^1 / 16^{cos^2 x}

Answer 7

then just put 16 ^{cos^2 x} = y

Answer 8

and u get a quadratic in y

Answer 9

thank you!

Answer 10

can u solve it now ?

Answer 11

first, by exponential rules,x^mx^n = x^(m+n)
so
16^(sin^2 x cos^2 x) = 10
natural log of both sides, pull out the exponent, divide by ln(16
sin^2 x times cos^2 x = ln10/ln16
sin^2cos^2 = (1/4)sin^2 (2x)
(1/4)sin^2 (2x) = ln10/ln16
multiply both sides by 4, square root both sides, inverse sine, and divide by 2

hope its right

Answer 12

nice one @hartnn

Answer 13

\[16^{\sin ^{2}x} = 2,8 \]?

Answer 14

yes.

Answer 15

now expressing 16 and 8 as powers of 2 and comparing the exponents

Answer 16

hmm... he went offline...

Answer 17

yes i got it
x= 30,60,330,300 (degrees)

Answer 18

those are correct values :)