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Math IIC Question

1104199411041994 Registered User Posts: 96 Junior Member
Solve
2sinx + cos2x = 2(sinx)^2 - 1

for 0≤x≤2pi

How do I solve this without a graphing calculator?
Post edited by 11041994 on

Replies to: Math IIC Question

  • rspencerspence Registered User Posts: 2,118 Senior Member
    cos 2x = 1 - 2 sin^2 (x), so we have

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

    Let u = sin x, this becomes a quadratic in terms of u. Solve for u. Remember that -1 <= u <= 1, since sin x can only range from -1 to 1, inclusive.
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