ss_6_7

# ss_6_7 - Section 6.7 Trigonometric Equations II In this...

This preview shows pages 1–2. Sign up to view the full content.

Section 6.7 Trigonometric Equations II In this section, we continue the study of trigonometric equations which was begun in the previous section. Example. Solve 2 cos 2 θ +cos θ - 1=0for0 θ 2 π . Solution. The quadratic formula gives cos θ = - 1 ± 1+8 4 = - 1 4 ± 3 4 = - 1 , 1 2 . cos θ = - 1 implies θ = π ± 2 . cos θ = 1 2 implies θ = π 3 ± 2 or θ = - π 3 ± 2 . The values of θ in [0 , 2 π ] satisfying the above are θ = π 3 , 5 π 3 . Example. Solve sin2 θ =cos θ on [0 , 2 π ]. Solution. sin 2 θ =2s in θ cos θ . Thus, sin 2 θ =co s θ iﬀ 2 sin θ cos θ =co s θ . Since cos θ =0i sa solution of 2 sin θ cos θ =cos θ and cos θ =0iﬀ θ = π 2 ± , θ = π 2 ± is a solution of sin 2 θ =cos θ . If cos θ 6 =0and2s in θ cos θ =cos θ ,then2sin θ =1ands in θ = 1 2 .I fs in θ = 1 2 ,then θ = π 6 ± 2 or θ

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## ss_6_7 - Section 6.7 Trigonometric Equations II In this...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online