7sin^2x-8sinxcosx=15cos^2x

Предыдущий
вопрос Следующий
вопрос

To solve this trigonometric equation, we can use the Pythagorean identity:

sin^2(x) + cos^2(x) = 1

Therefore, we can rewrite the equation as:

7sin^2(x) - 8sin(x)cos(x) = 15(1 - sin^2(x))

Expanding both sides, we get:

7sin^2(x) - 8sin(x)cos(x) = 15 - 15sin^2(x)

Rearranging terms, we get:

22sin^2(x) - 8sin(x)cos(x) - 15 = 0

Now, we can use trigonometric identities to rewrite the equation in terms of only sin(2x):

sin(2x) = 2sin(x)cos(x)

Therefore, our equation becomes:

22(sin(2x)/2)^2 - 8sin(2x)/2 - 15 = 0

Simplifying, we get:

11sin^2(2x) - 4sin(2x) - 15 = 0

Now, we can solve the quadratic equation for sin(2x) by factoring or using the quadratic formula. Once we find the values of sin(2x), we can solve for x by taking the arcsin of the values of sin(2x) and dividing by 2.