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Expresscos2xin terms ofcos2x, then solve the resulting quadratic to findcosx=1, hencex=2nπ... 更多結果 共享 已復制到剪貼板 示例 二次方程式 x2−4x−5=0 三角學 4sinθcosθ=2sinθ 線性方程 y=3x+4 算術 699∗533 矩陣
The 2cos(x+y2) terms in the numerator and denominator cancel out (assuming cos(x+y2)≠0): =sin(x−y2)cos(x−y2). Step 4: Recognize the tangent function The expression sin(x−y2)cos(x−y2) is the definition of the tangent function: =tan(x−y2). Conclusion Thus, ...
LHS=tan−1(cosx−sinxcosx+sinx) Step 2: Factor Out Common TermsNotice that we can factor out the terms in the numerator and denominator:LHS=tan−1(cosx−sinxcosx+sinx)=tan−1(1−tanx1+tanx)where tanx=sinxcosx. Step 3: Use the Tangent Subtraction FormulaWe can apply the ...
To find the values of p such that the equation 2cos2x−(p+3)cosx+2(p−1)=0 has a real solution, we can treat this as a quadratic equation in terms of cosx. Let's denote y=cosx. Then, we can rewrite the equation as: 2y2−(p+3)y+2(p−1)=0 Step 1: Identify the...