Solve 5sin^2x-1=0 ok

Question

asked Sep 8, 2022 141k views

1 Answer

Equasia · Answer 1 · 2022-09-14T07:17:17+0000

Answer:

to 1 decimal place,. (a) tan (3x + ... (i) 2cos 2x – 5sin 2x = 0 R cos(2x + λ) = 0 λ = 68.2 ⇒ 2x + 68.2 = 90 B1. 2x + λ = ... (16cos2x – 9)(cos2x – 1) = 0.

Solve 5sin^2x-1=0 ok

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

Please log in or register to add a comment.

Related questions

Categories

Other Questions