ss_6_7

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

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

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