document.write( "Question 979670: Find all solutions of the equation in the interval
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\n" ); document.write( "-sin2x+2cosx=0
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Algebra.Com's Answer #600961 by ikleyn(52778)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "The equation is\r
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\n" ); document.write( "\n" ); document.write( "\"2cosx\" = \"sin2x\".                      (1)\r
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\n" ); document.write( "\n" ); document.write( "Apply the formula for sines of the double argument:  \"sin2x\" = \"2sinx%2Acosx\"  (see,  for example,  the lesson  Trigonometric functions of multiply argument  in this site). \r
\n" ); document.write( "\n" ); document.write( "Then the equation takes the form\r
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\n" ); document.write( "\n" ); document.write( "\"2cosx\" = \"2%2Acosx%2Asinx\".             (2)\r
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\n" ); document.write( "\n" ); document.write( "One solution of this equation is  \"cosx\" = \"0\",  which gives  \"x\" = \"pi%2F2\",  \"3pi%2F2\"  (regarding the interval [0, 2pi) ).\r
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\n" ); document.write( "\n" ); document.write( "Further,  let us assume that  \"cosx\"  is not equal to zero.  Then we can reduce the equation  (2) dividing its both sides by  \"2%2Acosx\".  You will get\r
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\n" ); document.write( "\n" ); document.write( "\"1\" = \"sinx\", \r
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\n" ); document.write( "\n" ); document.write( "which gives the solutions  \"x\" = \"pi%2F2\"  we just obtained above. \r
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\n" ); document.write( "\n" ); document.write( "Answer.  The solutions are  \"x\" = \"pi%2F2\",  \"3pi%2F2\". \r
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